PHYS 202 - GENERAL PHYSICS II -
Kinematics in Two Dimensions
Question Bank - Set 4
Liberty University
Question 1
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. Calculate the maximum height reached by the
projectile.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity of the projectile can be resolved into two components: one
along the horizontal direction and the other along the vertical direction. The
horizontal component can be calculated as vix=vicos(θ), and the vertical
component can be calculated as viy=visin(θ). Given: - Initial speed, vi=
20 m/s - Launch angle, θ= 30◦
Therefore, we have: vix=vicos(θ) = 20 m/s ·cos(30◦)
viy=visin(θ) = 20 m/s ·sin(30◦)
Step 2: Calculate the time to reach maximum height. At the maximum
height, the vertical component of the projectile’s velocity is zero. The time taken
to reach the maximum height can be calculated using the vertical component of
the velocity and gravitational acceleration: vy=viy−gt. Given: - Acceleration
due to gravity, g= 9.81 m/s2
Setting vy= 0 and solving for t: 0 = viy−gt
Step 3: Calculate the maximum height reached by the projectile. The max-
imum height can be calculated using the equation: h=viy·t−1
2gt2. Substitute
the values of viyand tinto the equation to find the maximum height reached
by the projectile.
Question 2
Question
A soccer player kicks a ball from the corner of a field. The ball follows a parabolic
path and lands 35 meters from the player. The initial velocity of the ball is 20
m/s at an angle of 30◦above the horizontal. How high above the ground is the
ball kicked?
(Note: Assume the acceleration due to gravity is 9.81 m/s2.)
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity v0can be resolved into its horizontal component v0xand
vertical component v0yas follows:
v0x=v0cos θ= 20 m/s ·cos 30◦= 17.32 m/s
v0y=v0sin θ= 20 m/s ·sin 30◦= 10.00 m/s
Step 2: Determine the time of flight. Since the time of flight for an ob-
ject launched at an angle in a uniform gravity field is twice the time to reach
maximum height, we can find the time of flight as follows:
tflight = 2 ×v0y
g= 2 ×10.00 m/s
9.81 m/s2= 2.04 s
Step 3: Determine the maximum height. The maximum height hmax reached
by the ball can be calculated using the formula:
hmax =v2
0y
2g=(10.00 m/s)2
2×9.81 m/s2= 5.10 m
Step 4: Calculate the distance from the player to the point where the ball
lands. The distance dfrom the player to where the ball lands is given by:
d=v0x×tflight = 17.32 m/s ×2.04 s = 35.28 m
Step 5: Determine the height above the ground where the ball is kicked.
Since the ball lands 35 meters away, the height above the ground where the ball
is kicked can be calculated as the vertical displacement of the ball at that point:
hkicked =hmax+1
2gt2
flight = 5.10 m+0.5×9.81 m/s2×(2.04 s)2= 5.10 m+20.16 m = 25.26 m
Therefore, the ball is kicked 25.26 meters above the ground.
2
Question 3
Question
A particle moves along a curved path in the xy plane. At a certain instant, the
particle’s position vector r is given by r = (3t2−2)ˆ
i+ (4t+ 1)ˆ
j, where tis in
seconds and the position coordinates are in meters. Determine the particle’s
velocity and acceleration at t= 2 s.
Solution
Step 1: To find the particle’s velocity, we differentiate the position vector r with
respect to time t:
v =dr
dt
Step 2: Differentiating the components of r:
d
dt[(3t2−2)ˆ
i]=6tˆ
i
d
dt[(4t+ 1)ˆ
j]=4ˆ
j
Therefore, the velocity vector at any time tis given by:
v = 6tˆ
i+ 4ˆ
j
Step 3: Now, to find the particle’s acceleration, we differentiate the velocity
vector v with respect to time t:
a =dv
dt
Step 4: Differentiating the components of v:
d
dt[6tˆ
i]=6ˆ
i
d
dt[4ˆ
j] = 0
Therefore, the acceleration vector at any time tis given by:
a = 6ˆ
i
Step 5: Finally, we evaluate the velocity and acceleration vectors at t= 2 s:
v(2) = 6(2)ˆ
i+ 4ˆ
j= 12ˆ
i+ 4ˆ
jm/s
a(2) = 6ˆ
im/s2
Therefore, at t= 2 s, the particle’s velocity is 12ˆ
i+ 4ˆ
jm/s and the acceler-
ation is 6ˆ
im/s2.
3
Question 4
Question
A golf ball is hit with an initial velocity of 30 m/s at an angle of 45 degrees
above the horizontal. Ignoring air resistance, calculate the maximum height the
ball will reach and the total time it takes to reach the ground.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity of the golf ball can be resolved into horizontal and vertical
components as follows: The initial horizontal component of velocity v0xis given
by: v0x=v0·cos θwhere v0= 30 m/s and θ= 45◦. So, v0x= 30 ·cos 45◦.
The initial vertical component of velocity v0yis given by: v0y=v0·sin θ
where v0= 30 m/s and θ= 45◦. So, v0y= 30 ·sin 45◦.
Step 2: Calculate the time taken to reach the maximum height. The time
taken to reach the maximum height can be calculated using the vertical com-
ponent of motion. The equation to determine the time taken to reach the
maximum height is:
v0y=vy−gt
Where v0yis the initial vertical component of velocity, vyis the vertical compo-
nent of velocity at the maximum height (which will be zero m/s), g= 9.81 m/s2
is the acceleration due to gravity, and tis the time taken to reach the maximum
height.
Step 3: Calculate the maximum height the golf ball will reach. The max-
imum height reached by the golf ball can be calculated using the equation for
vertical displacement:
y=v0y·t−1
2gt2
where yis the maximum height, v0yis the initial vertical component of velocity,
tis the time taken to reach the maximum height, and gis the acceleration due
to gravity.
Step 4: Calculate the total time taken to reach the ground. The total time
taken by the golf ball to reach the ground can be calculated using the formula
for the time of flight in projectile motion:
T=2v0sin θ
g
where Tis the total time of flight, v0is the initial velocity, θis the angle of
projection, and gis the acceleration due to gravity.
4
Question 5
Question
A soccer player kicks a ball from the ground with an initial velocity of 20 m/s
at an angle of 35◦above the horizontal. The ball lands on the ground 80 meters
away. Calculate the maximum height reached by the ball during its flight.
Solution
Step 1: Split the initial velocity into its horizontal and vertical components. The
horizontal component of the initial velocity can be calculated as v0x=v0cos(θ),
where v0= 20 m/s and θ= 35◦.
v0x= 20 cos(35◦)≈16.45 m/s
The vertical component of the initial velocity can be calculated as v0y=
v0sin(θ).
v0y= 20 sin(35◦)≈11.47 m/s
Step 2: Determine the time of flight. The time of flight tcan be determined
using the vertical component of the motion. The equation of motion in the
vertical direction is y=v0yt−1
2gt2, where yis the maximum height reached by
the ball and g= 9.81 m/s2is the acceleration due to gravity. At the maximum
height, the vertical component of the velocity is zero. Therefore, v0y−gt = 0.
t=v0y
g=11.47
9.81 ≈1.17 s
Step 3: Calculate the maximum height reached by the ball. Substitute the
time of flight into the equation for vertical motion.
y=v0yt−1
2gt2= 11.47 ×1.17 −1
2×9.81 ×(1.17)2≈6.49 m
Therefore, the maximum height reached by the ball during its flight is ap-
proximately 6.49 meters.
Question 6
Question
A baseball player hits a ball with an initial velocity of 30 m/s at an angle of
30 degrees above the horizontal. What is the maximum height the ball reaches
during its flight? (Assume the ball is hit from ground level and air resistance is
negligible.)
5
Solution
Step 1: Resolve the initial velocity into horizontal and vertical components. The
horizontal component v0xis given by: v0x=v0cos(θ) where v0= 30 m/s is the
initial velocity and θ= 30◦is the angle. Therefore, v0x= 30 ×cos(30◦).
Step 2: Calculate the horizontal component of the initial velocity. v0x=
30 ×cos(30◦) = 30 ×√3
2= 15√3 m/s.
Step 3: Use the kinematic equation to find the time taken to reach the
maximum height. The vertical component v0yis given by: v0y=v0sin(θ).
Substitute v0= 30 m/s and θ= 30◦to find v0y.
Step 4: Calculate the vertical component of the initial velocity. v0y= 30 ×
sin(30◦) = 30 ×1
2= 15 m/s.
Step 5: Use the equation vf=vi+at for vertical motion to find the time
taken to reach the maximum height. Since the ball reaches its maximum height,
the final velocity at this point is 0. Therefore, 0 = v0y−gt.
Step 6: Solve for the time t.t=v0y
g=15
9.8.
Step 7: Calculate the time t.t≈1.53 s.
Step 8: Use the equation y=v0yt−1
2gt2to find the maximum height.
y= 15 ×1.53 −1
2×9.8×(1.53)2.
Step 9: Calculate the maximum height y.y≈22.95 m.
Therefore, the maximum height the ball reaches during its flight is approxi-
mately 22.95 meters.
Question 7
Question
A baseball pitcher throws a fastball with an initial velocity of 30 m/s at an angle
of 30◦above the horizontal. At the instant the baseball leaves the pitcher’s hand,
a batter standing at home plate 18 m away sees the ball. Will the ball pass over
the plate or will it hit the ground first? Assume the baseball is in projectile
motion and air resistance is negligible.
Solution
Step 1: Break the initial velocity vector into its horizontal and vertical compo-
nents. The horizontal component of the velocity is given by:
v0x=v0cos(θ) = 30 m/s ·cos(30◦)≈25.98 m/s
The vertical component of the velocity is given by:
v0y=v0sin(θ) = 30 m/s ·sin(30◦)≈15 m/s
Step 2: Determine the time it takes for the ball to reach the batter. Let’s
consider the vertical motion of the baseball. The vertical position of the ball at
6
time tis given by:
y=v0yt−1
2gt2
where g= 9.81 m/s2is the acceleration due to gravity.
Since the ball is at ground level when it reaches the batter, y= 0 and we
can solve for t:
0 = 15t−1
2·9.81 ·t2
t(15 −4.905t)=0
t= 0 s or t≈3.06 s
Step 3: Determine the horizontal distance the ball travels in this time. The
horizontal distance the ball travels is given by:
x=v0xt
x= 25.98 m/s ·3.06 s ≈79.49 m
Since the batter is only 18 m away, the ball will hit the ground before passing
over the plate.
Question 8
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. Find the maximum height above the ground
reached by the projectile. Take g= 9.8 m/s2.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity can be resolved into two components: v0x=v0cos(θ) and
v0y=v0sin(θ), where v0= 20 m/s and θ= 30◦. Thus, v0x= 20 m/s ·cos(30◦)
and v0y= 20 m/s ·sin(30◦).
Step 2: Find the time taken to reach the maximum height. Using vy=
v0y−gt and vy= 0 at the maximum height, we have 0 = 20 sin(30◦)−9.8t.
Solving for t, we get t=20 sin(30◦)
9.8.
Step 3: Calculate the maximum height. The maximum height hcan be
found using the formula h=ymax =v0yt−1
2gt2. Substitute the values of v0y
and tto find h. Remember that at the highest point vy= 0, so v0y−gt = 0.
Thus, h= 20 sin(30◦)·20 sin(30◦)
9.8−1
2·9.820 sin(30◦)
9.82. Simplify this expression
to find the maximum height.
7
Question 9
Question
A baseball is hit at an angle of 30◦above the horizontal with an initial speed of
40 m/s. At the same time, a gust of wind starts blowing the baseball horizontally
at a constant speed of 5 m/s. Calculate the maximum height above its starting
point that the baseball reaches.
Solution
Step 1: Resolve the initial velocity of the baseball into its horizontal and vertical
components. Let v0xbe the initial velocity in the x-direction and v0ybe the
initial velocity in the y-direction, then: v0x=v0cos(30◦) = 40 cos(30◦)≈34.64
m/s v0y=v0sin(30◦) = 40 sin(30◦)≈20 m/s
Step 2: Determine the time taken for the baseball to reach its maximum
height. The time taken to reach the maximum height can be calculated using
the equation vfy =v0y−gt, where vf y is the final velocity in the y-direction
(which is 0 m/s at the maximum height), gis the acceleration due to gravity
(approximately -9.81 m/s2), and tis the time.
0 = 20 −9.81t
t=20
9.81 ≈2.04 seconds
Step 3: Calculate the maximum height reached by the baseball. The maxi-
mum height can be determined by using the equation y=v0yt−1
2gt2, where y
is the height, v0yis the initial velocity in the y-direction, tis the time calculated
in step 2, and gis the acceleration due to gravity.
y= 20(2.04) −1
2(9.81)(2.04)2
y≈20.4 meters
Therefore, the baseball reaches a maximum height of approximately 20.4
meters above its starting point.
Question 10
Question
A baseball is hit with an initial velocity of 30 m/s at an angle of 45 degrees
above the horizontal. Find the maximum height the baseball reaches and the
total time it is in the air before hitting the ground. Assume air resistance is
negligible and take the acceleration due to gravity as 9.81 m/s2.
8
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity of the baseball can be resolved into its horizontal (v0x) and
vertical (v0y) components as follows:
v0x=v0cos θ= 30 cos 45◦≈21.21 m/s
v0y=v0sin θ= 30 sin 45◦≈21.21 m/s
Step 2: Find the time to reach maximum height. The time to reach the
maximum height (tpeak) can be calculated using the vertical component of the
initial velocity and the acceleration due to gravity:
vpeak = 0
vpeak =v0y−gtpeak
0 = 21.21 −9.81tpeak
tpeak =21.21
9.81 ≈2.16 s
Step 3: Calculate the maximum height. The maximum height (ymax) can
be found using the equation for vertical position:
ymax =v0ytpeak −1
2gt2
peak
ymax = 21.21 ×2.16 −1
2×9.81 ×(2.16)2
ymax ≈23.15 m
Step 4: Determine the total time in the air. The total time the baseball is
in the air can be calculated as twice the time to reach the peak:
Total time = 2tpeak = 2 ×2.16
Total time ≈4.32 s
Therefore, the baseball reaches a maximum height of approximately 23.15
m and is in the air for approximately 4.32 seconds before hitting the ground.
Question 11
Question
A ball is thrown from the edge of a cliff at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. The cliff is 50 m high. How far from the base
of the cliff will the ball hit the ground?
9
Solution
Let’s break down the motion of the ball into its horizontal and vertical compo-
nents. We can then solve for the time it takes for the ball to hit the ground and
use that time to find the horizontal distance traveled.
Step 1: Find the time it takes for the ball to hit the ground.
We know the initial vertical velocity of the ball is 20 sin(30◦) m/s.
The acceleration due to gravity is −9.8 m/s2.
The ball will hit the ground when its vertical position is 0.
Using the equation of motion:
yf=yi+viyt+1
2at2
we can set yf= 0, yi= 50 m, viy= 20 sin(30◦) m/s, and a=−9.8 m/s2to
solve for t.
0 = 50 + 20 sin(30◦)t−4.9t2
Solving this quadratic equation gives t≈3.25 s.
Step 2: Find the horizontal distance traveled by the ball.
The initial horizontal velocity of the ball is 20 cos(30◦) m/s.
The horizontal distance the ball travels is given by:
d=vixt
Substitute vix= 20 cos(30◦) m/s and t≈3.25 s to find:
d≈(20 cos(30◦)·3.25) m
Calculating this value, we find that the ball will hit the ground approximately
56.4 m from the base of the cliff.
Question 12
Question
A soccer player kicks a ball with an initial velocity of 20 m/s at an angle of 30
degrees above the horizontal. How far does the ball travel horizontally before it
hits the ground? (Assume no air resistance and g= 9.8 m/s2)
10
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity can be broken down into its horizontal and vertical com-
ponents using trigonometric functions. The horizontal component is given by
v0x=v0cos(θ) and the vertical component is v0y=v0sin(θ), where v0= 20 m/s
and θ= 30◦.
Step 2: Calculate the time of flight. The time taken for the ball to reach the
ground can be determined using the vertical component of the velocity. The
time of flight tcan be found using the equation y=v0yt+1
2gt2, where y= 0
(since the ball starts and ends at ground level).
Step 3: Calculate the horizontal distance traveled. The horizontal distance
traveled by the ball can be calculated using the horizontal component of the
velocity and the time of flight. The horizontal distance xis given by x=v0xt.
Step 4: Substitute the values and solve for the horizontal distance. Substi-
tute the given values (v0= 20 m/s, θ= 30◦,g= 9.8 m/s2) into the equations
derived in Steps 1 to 3, and solve for the horizontal distance the ball travels
before hitting the ground.
Question 13
Question
A particle moves in the xy plane according to the following set of equations:
x= (3.00 m/s)tcos(30◦)
y= (3.00 m/s)tsin(30◦)−1
2(9.80 m/s2)t2
At what time does the particle cross the xaxis for the first time?
Solution
Step 1: Find the xcoordinate of the particle crossing the xaxis for the first
time. The particle crosses the xaxis when y= 0, so we substitute y= 0 into
the given equations:
0 = (3.00 m/s)tsin(30◦)−1
2(9.80 m/s2)t2
0 = (3.00 m/s)t1
2−1
2(9.80 m/s2)t2
0 = 1.50t−4.90t2
Step 2: Solve the quadratic equation 1.50t−4.90t2= 0. Factoring out t, we
get:
t(1.50 −4.90t)=0
11
Thus, t= 0 or t=1.50
4.90 ≈0.31 s.
Step 3: Since the particle is already moving in the positive xdirection at
t= 0, we can conclude that the particle crosses the xaxis for the first time at
t≈0.31 s.
Question 14
Question
A soccer player kicks a ball from the ground with an initial velocity of 20 m/s
at an angle of 30 degrees above the horizontal. The ball lands on the ground
60 meters away. Calculate: a) The time of flight of the ball. b) The maximum
height the ball reaches. c) The magnitude of the ball’s velocity just before it
hits the ground.
Solution
a) To find the time of flight, we can use the equation for the horizontal displace-
ment of a projectile:
∆x=Vix·t
where ∆xis the horizontal displacement, Vixis the initial horizontal velocity,
and tis the time of flight.
Step 1: Calculate the initial horizontal velocity component:
Vix=Vi·cos θ
Vix= 20 m/s ·cos(30◦)
Vix= 20 m/s ·√3
2
Vix= 10√3 m/s
Step 2: Substitute the known values into the equation and solve for t:
60 = 10√3 m/s ·t
t=60
10√3
t=6√3
3
t= 2√3 seconds
Therefore, the time of flight of the ball is 2√3 seconds.
12
b) To find the maximum height, we can use the vertical component of the
initial velocity and the time of flight along with the equation for vertical dis-
placement of a projectile:
∆y=Viy·t+1
2·a·t2
where: - ∆yis the vertical displacement, - Viyis the initial vertical velocity
component, - tis the time of flight, and - ais the acceleration due to gravity.
Step 3: Calculate the initial vertical velocity component:
Viy=Vi·sin θ
Viy= 20 m/s ·sin(30◦)
Viy= 20 m/s ·1
2
Viy= 10 m/s
Step 4: Substitute the known values into the equation and solve for the
maximum height:
∆y= 10 m/s ·2√3 + 1
2·(−9.8 m/s2)·(2√3)2
∆y= 20√3−29.4×3
∆y= 20√3−88.2
∆y≈3.188 meters
Therefore, the maximum height the ball reaches is approximately 3.188 me-
ters.
c) The magnitude of the ball’s velocity just before it hits the ground is equal
to the initial speed since no horizontal force acts on the ball.
Hence, the magnitude of the ball’s velocity just before it hits the ground is
20 m/s.
Question 15
Question
A projectile is launched from the ground at an angle of 45◦above the horizontal
with an initial speed of 20 m/s. Find the total time the projectile is in the
air and the maximum height it reaches. Your answer should include both the
mathematical expressions and the numerical values.
13
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity v0can be resolved into horizontal and vertical components
as follows:
v0x=v0cos(θ) and v0y=v0sin(θ)
where v0xis the initial velocity in the x-direction and v0yis the initial velocity
in the y-direction. Here, v0= 20 m/s and θ= 45◦.
Step 2: Calculate the initial velocity components v0xand v0y. Substitute
v0= 20 m/s and θ= 45◦into the equations:
v0x= 20 cos(45◦) = 20 ×√2
2= 10√2 m/s
v0y= 20 sin(45◦) = 20 ×√2
2= 10√2 m/s
Step 3: Find the time of flight. The time of flight can be determined by
using the y-component of the displacement equation:
y=v0yt−1
2gt2
Since the projectile returns to the same height it was launched from, y= 0 and
solving for tgives:
0 = 10√2t−1
2(9.8)t2
Solving for twe get two solutions: t= 0 or t≈2.03 s. Since t= 0 is the initial
time, the total time the projectile is in the air is approximately 2.03 seconds.
Step 4: Find the maximum height. The maximum height can be determined
by using the y-component of the displacement equation:
y=v0yt−1
2gt2
at the top of its trajectory, the vertical component of velocity is 0. Therefore,
v0y−gt = 0.
t=v0y
g=10√2
9.8≈1.43 s
Substitute t= 1.43 s into the equation gives:
y= 10√2×1.43 −1
2×9.8×(1.43)2≈14.1 m
Thus, the total time the projectile is in the air is approximately 2.03 seconds
and the maximum height it reaches is approximately 14.1 meters.
14
Question 16
Question
A baseball is hit with an initial velocity of 30 m/s at an angle of 40◦above the
horizontal. Calculate the maximum height the baseball reaches and the total
horizontal distance it travels before hitting the ground. Assume air resistance
is negligible and the acceleration due to gravity is 9.81 m/s2.
Solution
Step 1: Break the initial velocity into its horizontal and vertical components.
The horizontal component of velocity, v0x, can be calculated as:
v0x=v0cos(θ)
v0x= 30 cos(40◦)
v0x≈23.05 m/s
The vertical component of velocity, v0y, can be calculated as:
v0y=v0sin(θ)
v0y= 30 sin(40◦)
v0y≈19.34 m/s
Step 2: Calculate the time taken to reach maximum height. The time taken
to reach maximum height can be calculated using the vertical component of
velocity and acceleration due to gravity:
vfy =v0y−gt
0 = 19.34 −9.81t
t=19.34
9.81
t≈1.97 s
Step 3: Calculate the maximum height reached by the baseball. The max-
imum height can be calculated using the vertical component of velocity and
time:
ymax =v0yt−1
2gt2
ymax = 19.34 ·1.97 −1
2·9.81 ·(1.97)2
ymax ≈19.08 m
15
Step 4: Calculate the total horizontal distance traveled by the baseball. The
total horizontal distance can be calculated using the horizontal component of
velocity and time:
xtotal =v0x·t
xtotal = 23.05 ·1.97
xtotal ≈45.34 m
Therefore, the baseball reaches a maximum height of approximately 19.08 m
and travels a total horizontal distance of approximately 45.34 m.
Question 17
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. Find the total time of flight, the maximum
height reached, and the horizontal range of the projectile. Assume the acceler-
ation due to gravity is −9.81 m/s2.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity of the projectile can be resolved into horizontal (v0x) and
vertical (v0y) components as:
v0x=v0cos(θ)
v0y=v0sin(θ)
Given that v0= 20 m/s and θ= 30◦, we can calculate v0xand v0yas follows:
v0x= 20 ×cos(30◦) = 20 ×√3
2= 10√3 m/s
v0y= 20 ×sin(30◦) = 20 ×1
2= 10 m/s
Step 2: Find the time of flight. The total time of flight (T) can be found
using the y-component of the motion. At the maximum height, the vertical
component of the velocity is zero. The time taken to reach the maximum height
is equal to the time taken for the projectile to descend from this point to the
ground.
Using the equation vy=v0y−gt, where vy= 0 at the maximum height, we
can find the time taken to reach the maximum height:
0 = 10 −9.81t
t=10
9.81 = 1.02 s
16
Therefore, the total time of flight is twice the time taken to reach the max-
imum height:
T= 2t= 2 ×1.02 = 2.04 s
Step 3: Find the maximum height reached. To find the maximum height
(hmax) reached by the projectile, we use the equation:
hmax =v2
0y/(2g)
Substitute the known values to find hmax:
hmax = (10)2/(2 ×9.81) = 5.1 m
Step 4: Find the horizontal range. The horizontal range of the projectile is
the horizontal component of the velocity multiplied by the total time of flight:
R=v0x·T= 10√3·2.04 ≈20.8 m
Therefore, the total time of flight is 2.04 s, the maximum height reached is
5.1 m, and the horizontal range of the projectile is approximately 20.8 m.
Question 18
Question
A car travels along a straight road with an initial velocity of 20 m/s. It acceler-
ates at a constant rate of 2 m/s2for 5 seconds, and then continues at a constant
velocity for another 10 seconds. At this point, the driver sees an obstacle ahead
and applies the brakes, causing the car to decelerate at a rate of 3 m/s2. Find
the total distance the car travels during the entire trip.
Solution
Step 1: Find the distance traveled during acceleration phase. The distance d1
traveled during the acceleration phase can be found using the equation d1=
vit+1
2at2, where viis the initial velocity, ais the acceleration, and tis the
time. Plugging in the values, we get: d1= (20 m/s)(5 s) + 1
2(2 m/s2)(5 s)2
d1= 100 m + 25 m d1= 125 m
Step 2: Find the distance traveled during constant velocity phase. The dis-
tance d2traveled during the constant velocity phase can be found by multiplying
the velocity by time. d2= (20 m/s)(10 s) d2= 200 m
Step 3: Find the distance traveled during deceleration phase. The distance
d3traveled during the deceleration phase can be found using the equation d3=
vft−1
2at2, where vfis the final velocity, ais the deceleration, and tis the
time. First, calculate the final velocity before braking: Vf=Vi+at = 20 m/s +
2 m/s2·5 s = 30 m/s.Then, calculate the distance traveled during deceleration:
d3= (30 m/s)(10 s) −1
2(3 m/s2)(10 s)2d3= 300 m −150 m d3= 150 m
17
Step 4: Find the total distance traveled. The total distance traveled is the
sum of the distances traveled during each phase: Total distance = d1+d2+d3=
125 m + 200 m + 150 m = 475 m
Therefore, the total distance the car travels during the entire trip is 475
meters.
Question 19
Question
A car travels along a straight road and then around a curve that is part of a
circular racetrack. The car covers the second half of the curve in half the time
it covers the first half. If the total distance covered is 1.5 km and the average
speed is 60 km/h, what is the radius of the curve?
Solution
Step 1: Let’s denote the distance covered in the first half of the curve as d1and
the distance covered in the second half as d2. Since the car’s average speed is
60 km/h, we can write:
d1+d2= 1.5 km
The time taken to cover the first half, t1, is twice the time taken to cover the
second half, t2. We can write:
t1= 2t2
Step 2: The average speed of the car is given by:
Average speed = total distance
total time =d1+d2
t1+t2
Step 3: We can rewrite the average speed using the relationship between
distance, time, and speed:
Average speed = d1+d2
t1+t2
=d1+d2
d1
v+d2
2v
where vis the speed of the car.
Step 4: Substituting the known values into the equation:
60 = 1.5
d1
60 +d1
120
Step 5: Solving for d1:
60 = 1.5
3
2d1
d1= 1 km
18
Step 6: Substituting d1back into the total distance equation:
1 + d2= 1.5
d2= 0.5 km
Step 7: The time taken to cover the first half is:
t1=1
60 =1
60 hour
Step 8: The time taken to cover the second half is:
t2=0.5
60 =1
120 hour
Step 9: Using the formula for centripetal acceleration, a=v2
r, where vis
the speed of the car and ris the radius of the curve, we have:
a=(60 km/h)2
r
Step 10: At the midway point of the curve, the car’s acceleration must be
equal to the acceleration due to gravity to keep the car moving along the curve.
Thus,
(60 km/h)2
r=g
where g= 9.81 m/s2is the acceleration due to gravity.
Step 11: Converting the speed to m/s:
60 km/h = 60 ×1000
3600 m/s = 500
3m/s
Step 12: Substituting the values into the equation:
(500
3)2
r= 9.81
r=(500
3)2
9.81
r≈278.3 m
Therefore, the radius of the curve is approximately 278.3 m.
Question 20
Question
A baseball player hits a ball that is caught by an outfielder 3.5 seconds later. If
the ball was hit at an angle of 30◦above the horizontal and was caught at the
same height it was hit, what was the initial velocity of the ball?
19
Solution
Step 1: Identify known variables and convert angle to radians. Step 2: Break
initial velocity into horizontal and vertical components. Step 3: Use the vertical
motion equation to find the time the ball spends in the air. Step 4: Use the
horizontal motion equation to find the initial velocity. Step 5: Calculate the
initial velocity of the ball.
Step 1: The known variables are time t= 3.5 s, angle θ= 30◦, and accel-
eration due to gravity g= 9.8 m/s2. Converting the angle to radians, we have
θ= 30◦×π
180 =π
6radians.
Step 2: The initial velocity v0can be broken down into horizontal v0xand
vertical v0ycomponents:
v0x=v0cos(θ) and v0y=v0sin(θ)
Step 3: In the vertical direction, the motion equation is:
y=v0yt−1
2gt2
Since the ball was caught at the same height it was hit, the height is 0, and the
equation simplifies to:
0 = v0sin(θ)t−1
2gt2
Solving for t, we get:
t=2v0sin(θ)
g
Step 4: In the horizontal direction, the motion equation is:
x=v0xt
Substitute v0x=v0cos(θ) and t=2v0sin(θ)
g:
x=v0cos(θ)·2v0sin(θ)
g
Step 5: The initial velocity of the ball can be found by solving for v0:
x=2v2
0sin(θ) cos(θ)
g
v2
0=gx
2 sin(θ) cos(θ)
v0=rgx
2 sin(θ) cos(θ)
Substitute x= 0 (initial position) and θ=π
6:
v0=s9.8×0
2 sin π
6cos π
6= 0
Therefore, the initial velocity of the ball was 0 m/s.
20
Question 21
Question
A tennis ball is hit at an angle of 30◦above the horizontal from a height of 2
meters at an initial speed of 20 m/s. How far from the base of the wall does the
ball land if the wall is 10 meters high?
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical compo-
nents. The initial velocity can be broken down into its horizontal and verti-
cal components as follows: Vix=Vicos θ= 20 m/s ×cos(30◦) = 17.32 m/s
Viy=Visin θ= 20 m/s ×sin(30◦) = 10 m/s
Step 2: Determine the time taken to reach the top of the wall. We can
calculate the time taken to reach the highest point using the vertical component
of the initial velocity. The equation we can use is: Vf=Vi+at Where Vfis the
final vertical velocity (0 m/s at the top), ais the acceleration due to gravity (-9.8
m/s2), and tis the time taken. Substitute the known values into the equation
and solve for t: 0 = 10 −9.8t t =10
9.8= 1.02 s
Step 3: Calculate the horizontal distance the ball travels in the air. The
horizontal distance the ball travels is given by the equation: d=Vix×tSubsti-
tute the values of Vixand tinto the equation: d= 17.32 m/s ×1.02 s = 17.66
m.
Step 4: Determine the height reached by the ball at the point it lands.
The height reached by the ball when it lands can be calculated using the ver-
tical component of the initial velocity and the time taken to hit the wall. The
equation we can use is: H=Viyt−1
2gt2Substitute the known values into the
equation and solve for H:H= 10 ×1.02 −1
2×9.8×(1.02)2= 5.1−5 = 0.1 m
Step 5: Determine the distance from the base of the wall where the ball lands.
The horizontal distance from the wall where the ball lands is the sum of the dis-
tance traveled horizontally and the distance it fell vertically. Given that the wall
is 10 meters high, the total distance is: 17.66 m+p(10 m −2 m)2+ (0.1 m)2=
17.66 m + √64.1≈25.33 m
Therefore, the ball lands approximately 25.33 meters from the base of the
wall.
Question 22
Question
A projectile is launched at an angle of 30◦above the horizontal with an initial
speed of 20 m/s. Calculate the time it takes for the projectile to reach its
maximum height. Assume that air resistance is negligible and take g= 9.8
m/s2.
21
Solution
Step 1: Resolve the initial velocity into its xand ycomponents. The initial
velocity is given as 20 m/s at an angle of 30◦above the horizontal. Resolving
this into its xand ycomponents, we get: V0x= 20 cos 30◦= 20 ·√3
2= 10√3
m/s, V0y= 20 sin 30◦= 20 ·1
2= 10 m/s.
Step 2: Calculate the time it takes to reach maximum height. At the highest
point, the vertical component of velocity becomes 0. We can use the equation:
Vy=V0y−gt, 0 = 10 −9.8t,t=10
9.8≈1.02 seconds.
Therefore, it takes approximately 1.02 seconds for the projectile to reach its
maximum height.
Question 23
Question
A football is kicked with an initial velocity of 20 m/s at an angle of 30◦above
the horizontal. Find the maximum height the football reaches and the time it
takes to reach this height.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
Let v0= 20 m/s be the initial velocity and θ= 30◦be the angle above the
horizontal.
The initial horizontal velocity v0xis given by v0x=v0cos θ.
The initial vertical velocity v0yis given by v0y=v0sin θ.
Step 2: Determine the time it takes to reach the maximum height.
The time it takes to reach the maximum height is given by the vertical
motion equation y=v0yt−1
2gt2, where y= 0 (maximum height) and
g= 9.81 m/s2is the acceleration due to gravity.
Substitute v0yand ginto the equation and solve for t.
Step 3: Find the maximum height.
The maximum height is given by the equation ymax =v0yt−1
2gt2using
the time tobtained in step 2.
Substitute v0y,t, and ginto the equation and calculate ymax.
22
Question 24
Question
A particle moves in the xy-plane with an acceleration given by a = 4tˆ
i+ 3ˆ
j
m/s2, where ˆ
iand ˆ
jare unit vectors in the xand ydirections, respectively. At
t= 0, the particle is at the origin with a velocity of 5ˆ
im/s. Find the speed of
the particle when it reaches a position 10 m along the x-axis at an angle of 30◦
above it.
Solution
Step 1: The components of velocity are obtained by integrating the components
of acceleration with respect to time.
Zaxdt =Z4t dt = 2t2+C1
Zaydt =Z3dt = 3t+C2
Given that at t= 0, the particle has a velocity of 5ˆ
im/s, we have:
vx(0) = 5ˆ
i= 0ˆ
i+C1⇒C1= 5ˆ
i
vy(0) = 0ˆ
j+C2= 0ˆ
j⇒C2= 0ˆ
j
Thus, the velocity components are:
vx= 2t2+ 5
vy= 3t
Step 2: We can find speed from the velocity components.
v=qv2
x+v2
y
v(t) = p(2t2+ 5)2+ (3t)2
At the final position, x= 10 m and y= 10 tan 30◦= 5√3 m. We need to find
the time tat which x= 10 m by solving x=Rvxdt =R(2t2+ 5) dt.
10 = Z(2t2+ 5) dt =2
3t3+ 5t+C3
Given initial velocity vx(0) = 5, we can evaluate the constant C3:
5=03+ 5 ∗0 + C3⇒C3= 5
Thus, x(t) = 2
3t3+ 5t+ 5.
23
Step 3: To find the time tat which x= 10 m, we solve for tin the equation
x(t) = 10.
10 = 2
3t3+ 5t+ 5
2
3t3+ 5t−5 = 0
This cubic equation must be solved numerically. Once tis found and substituted
back into the expression for v(t), the speed of the particle when it reaches the
specified position can be calculated.
Question 25
Question
A football is kicked at an angle of 30◦above the horizontal with an initial
speed of 20 m/s. How far does the football travel horizontally before hitting the
ground? (Assume no air resistance)
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity v0can be resolved into two components: v0x(horizontal
component) and v0y(vertical component).
v0x=v0cos(θ) = (20 m/s) cos(30◦) = 17.32 m/s
v0y=v0sin(θ) = (20 m/s) sin(30◦) = 10 m/s
Step 2: Determine the time of flight. The time taken for the football to
reach the ground can be determined using the vertical component of motion.
We will use vyf =v0y+at and y=v0yt+1
2at2. When the football hits the
ground, y= 0 and vyf = 0.
0 = 10 m/s + (−9.8 m/s2)t
t=10 m/s
9.8 m/s2= 1.02 s
Step 3: Calculate the horizontal distance traveled. The horizontal distance
xtraveled by the football can be calculated using the formula x=v0xt.
x= (17.32 m/s)(1.02 s) = 17.66 m
Therefore, the football travels approximately 17.66 meters horizontally be-
fore hitting the ground.
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Question 26
Question
A projectile is launched from the ground with an initial speed of 30 m/s at an
angle of 30◦above the horizontal. Find the maximum height reached by the
projectile.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The horizontal component is v0x=v0cos(θ) and the vertical component is
v0y=v0sin(θ). Step 2: Calculate the time it takes for the projectile to reach
maximum height using the equation vf=vi+a∆t, where vf= 0 at maximum
height. Step 3: Use the time found in Step 2 to calculate the maximum height
using the formula ymax =yi+v0yt−1
2gt2, where yi= 0 and g= 9.81 m/s2is
the acceleration due to gravity.
Solution
Step 1: The initial velocity can be resolved into its horizontal and vertical
components:
v0x=v0cos(θ) = 30 m/s ×cos(30◦) = 30√3/2 m/s
v0y=v0sin(θ) = 30 m/s ×sin(30◦) = 15 m/s
Step 2: The time to reach maximum height can be found using the vertical
component of velocity:
vf=vi+a∆t
0 = 15 m/s −9.81 m/s2×t
t=15 m/s
9.81 m/s2
t≈1.53 s
Step 3: The maximum height can be found using the vertical motion equa-
tion:
ymax =yi+v0yt−1
2gt2
ymax = 0 + 15 m/s ×1.53 s −1
2×9.81 m/s2×(1.53 s)2
ymax ≈11.5 m
Therefore, the maximum height reached by the projectile is approximately
11.5 meters.
25
Question 27
Question
A particle moves along a curved path in the xy plane. The position of the
particle as a function of time is given by the equations:
x(t)=3t2−4t
y(t)=2t3−6t2
Find the velocity and acceleration vectors of the particle at time t= 2.
Solution
Step 1: Find the velocity vector by taking the derivative of the position vector
with respect to time.
Velocity, v(t) = dr
dt =d
dt(x(t)ˆ
i+y(t)ˆ
j)
v(t) = d
dt(3t2−4t)ˆ
i+d
dt(2t3−6t2)ˆ
j
v(t) = (6t−4) ˆ
i+ (6t2−12t)ˆ
j
Step 2: Find the acceleration vector by taking the derivative of the velocity
vector with respect to time.
Acceleration, a(t) = dv
dt =d
dt((6t−4) ˆ
i+ (6t2−12t)ˆ
j)
a(t) = (6)ˆ
i+ (12t−12)ˆ
j
Step 3: Evaluate the velocity and acceleration vectors at t= 2.
At t= 2 :
v(2) = (6(2) −4) ˆ
i+ (6(2)2−12(2)) ˆ
j= 8 ˆ
i+ 12 ˆ
j
a(2) = (6) ˆ
i+ (12(2) −12) ˆ
j= 6 ˆ
i+ 12 ˆ
j
Therefore, at t= 2, the velocity vector is 8 ˆ
i+ 12 ˆ
jand the acceleration
vector is 6 ˆ
i+ 12 ˆ
j.
Question 28
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 40 m/s. Determine: (a) the maximum height above the
ground reached by the projectile, (b) the total time it takes for the projectile
to return to the ground.
26
Solution
Let’s solve the problem step by step:
Step 1: Resolve the initial velocity into its horizontal and vertical
components. The initial velocity of the projectile can be resolved into hori-
zontal and vertical components as follows: The initial velocity in the x-direction
(v0x) is given by:
v0x=v0cos(θ)
where v0= 40 m/s is the initial speed and θ= 30◦is the launch angle. Substitute
v0= 40 m/s and θ= 30◦into the equation:
v0x= 40 m/s ·cos(30◦) = 34.64 m/s
The initial velocity in the y-direction (v0y) is given by:
v0y=v0sin(θ)
Substitute v0= 40 m/s and θ= 30◦into the equation:
v0y= 40 m/s ·sin(30◦) = 20 m/s
Step 2: Calculate the time to reach maximum height (tmax). The
time taken to reach maximum height can be found using the equation:
vmax =v0y−g·tmax
where vmax = 0 m/s at the maximum height and g= 9.8 m/s2is the acceleration
due to gravity. Substitute vmax = 0 m/s, v0y= 20 m/s, and g= 9.8 m/s2into
the equation:
0 = 20 m/s −9.8 m/s2·tmax
Solving for tmax:
tmax =20 m/s
9.8 m/s2= 2.04 s
Step 3: Calculate the maximum height above the ground. The
maximum height reached by the projectile can be found using the equation:
ymax =v0y·tmax −1
2·g·(tmax)2
Substitute v0y= 20 m/s, tmax = 2.04 s, and g= 9.8 m/s2into the equation:
ymax = 20 m/s ·2.04 s −1
2·9.8 m/s2·(2.04 s)2
Calculating ymax:
ymax = 20 m/s ·2.04 s −1
2·9.8 m/s2·(2.04 s)2= 20.4 m
Step 4: Calculate the total time of flight. The total time of flight can
be determined by finding the time it takes for the projectile to return to the
ground, which is twice
27
Question 29
Question
A particle moves in the xy plane such that its position vector is given by r =
(4t2−2t)i+ (2t3+t)j, where tis in seconds. Determine the velocity and
acceleration vectors of the particle when t= 2 s.
Solution
Step 1: To find the velocity vector, we need to differentiate the position vector
r with respect to time t.
v =dr
dt
Step 2: Differentiating each component of the position vector r:
v =d
dt[(4t2−2t)i+ (2t3+t)j]
v = (8t−2)i+ (6t2+ 1)j
Step 3: Now, we need to find the acceleration vector by differentiating the
velocity vector v with respect to time t.
a =dv
dt
Step 4: Differentiating each component of the velocity vector v:
a =d
dt[(8t−2)i+ (6t2+ 1)j]
a = 8i+ 12tj
Step 5: Now, substitute t= 2 s into the velocity and acceleration vectors to
find their values at t= 2 s.
vt=2 = (8(2) −2)i+ (6(2)2+ 1)j
vt=2 = 14i+ 25j
at=2 = 8i+ 12(2)j
at=2 = 8i+ 24j
Therefore, the velocity vector at t= 2 s is vt=2 = 14i+ 25jm/s, and the
acceleration vector at t= 2 s is at=2 = 8i+ 24jm/s
²
.
28
Question 30
Question
A particle moves in the xy plane such that its position vector at time tis given
by r(t) = (3t2+ 2t)ˆ
i−(4t+ 1)ˆ
j, where ˆ
iand ˆ
jare unit vectors in the xand
ydirections, respectively. Find the magnitude of the particle’s acceleration at
t= 2 s.
Solution
Step 1: Find the particle’s velocity at t= 2 s by differentiating the position
vector:
v(t) = dr
dt = (6t+ 2)ˆ
i−4ˆ
j
Substitute t= 2 s to find the velocity at t= 2 s:
v(2) = (6(2) + 2)ˆ
i−4ˆ
j= 14ˆ
i−4ˆ
j
Step 2: Find the particle’s acceleration at t= 2 s by differentiating the
velocity vector:
a(t) = dv
dt = 6ˆ
i
Therefore, the magnitude of the particle’s acceleration at t= 2 s is:
|a(2)|=|6ˆ
i|= 6 m/s2
Question 31
Question
A particle is initially at the origin at t= 0. The particle then moves in the
xy-plane and has a position vector r= (4t3−2t)i+ (6t2+ 4)j, where iand jare
unit vectors in the xand ydirections, respectively. Find the magnitude of the
particle’s acceleration vector at t= 2 seconds.
Solution
Step 1: Find the velocity vector v(t) by taking the derivative of the position
vector r(t) with respect to time:
v(t) =
dt = ddt[(4t3−2t)i+(6t2+4)j]
v(t) = (12t2−2)i+ (12t)j
29
Step 2: Find the acceleration vector a(t) by taking the derivative of the
velocity vector v(t) with respect to time:
a(t) =
dt = ddt[(12t2−2)i+(12t)j]
a(t) = (24t)i+ 12j
Step 3: Find the acceleration vector aat t= 2 seconds:
a= (24(2))i+ 12j
a= 48i+ 12j
Step 4: Calculate the magnitude of the acceleration vector:
|a|=p(48)2+ (12)2
|a|=√2304 + 144
|a|=√2448
|a| ≈ √2400
|a| ≈ 49.38 m/s2
Therefore, the magnitude of the particle’s acceleration vector at t= 2 sec-
onds is approximately 49.38 m/s2.
Question 32
Question
A projectile is launched with an initial speed of 30 m/s at an angle of 30◦above
the horizontal from the ground. Determine the time taken for the projectile to
reach the maximum height. (Use g= 9.81 m/s2for acceleration due to gravity)
Solution
Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial velocity (vi) can be resolved into horizontal (vix) and vertical (viy)
components using trigonometry:
vix =vi·cos(θ)
viy =vi·sin(θ)
where vi= 30 m/s and θ= 30◦.
Step 2: Calculate the time taken to reach the maximum height. At the peak
of the projectile’s trajectory, the vertical velocity component is zero. We can
30
use this information to find the time taken to reach the maximum height. The
vertical velocity at any time tis given by:
vfy =viy −gt
At the maximum height, vfy = 0:
0 = viy −gt
t=viy
g
Substitute viy and gto find the time taken to reach the maximum height.
Step 3: Substitute the known values and calculate the time. Calculate vix,
viy, and substitute them into the equation for time:
vix = 30 ·cos(30◦) = 25.98 m/s
viy = 30 ·sin(30◦) = 15 m/s
Now substitute viy into the equation:
t=15
9.81 = 1.53 seconds
Therefore, the time taken for the projectile to reach the maximum height is
1.53 seconds.
Question 33
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. At the highest point of its trajectory, the
projectile explodes into two fragments of equal mass. One fragment immediately
drops vertically, and the other fragment continues to move along the original
path after the explosion. What is the speed of the fragment that continues to
move along the original path just after the explosion?
Solution
Step 1: Resolve initial velocity into horizontal and vertical components: The
initial velocity of the projectile can be resolved into its horizontal and vertical
components as: Horizontal component: V0x=V0·cos(30◦) Vertical component:
V0y=V0·sin(30◦)
Given that V0= 20 m/s, we substitute to find: V0x= 20·cos(30◦) = 20·√3
2=
10√3 m/s V0y= 20 ·sin(30◦) = 20 ·1
2= 10 m/s
Step 2: Time to reach highest point: The time taken for the projectile to
reach the highest point can be found using the vertical component of velocity.
31
At the highest point, the vertical component of velocity becomes 0. We can use
this information to find the time taken. The vertical component of velocity can
be given by: Vy=V0y−gt where Vy= 0 at the highest point.
Substitute V0y= 10 m/s and g= 9.81 m/s2to solve for t: 0 = 10 −9.81t
t=10
9.81 ≈1.02 s
Step 3: Time of flight for the projectile: Since the projectile reaches highest
point after time t, the total time of flight will be double this time. Therefore,
the time of flight for the projectile is 2t= 2 ·1.02 = 2.04 s.
Step 4: Horizontal distance traveled by the projectile: The horizontal dis-
tance traveled by the projectile can be found using the horizontal compo-
nent of velocity and the time of flight. The horizontal distance is given by:
X=V0x·tSubstitute V0x= 10√3 m/s and t= 2.04 s to solve for X:
X= 10√3·2.04 = 20.4√3 meters
Step 5: Speed of the fragment moving along the original path after explosion:
Since the vertical component of velocity of the projectile immediately drops to
zero after explosion, the horizontal component will remain unchanged. Hence,
the speed of the fragment moving along the original path after the explosion is
equal to the horizontal component of velocity, which is 10√3 m/s.
Question 34
Question
A particle moves in the xy plane such that its position is given by r(t) =
(4t2+ 2t)i+ (3t−t3)j, where tis in seconds and the position components are
in meters. Find the magnitude of the velocity of the particle at t= 2 s.
Solution
Step 1: Find the velocity function v(t) by taking the derivative of the position
function r(t).
Step 1: v(t) = dr(t)
dt =d(4t2+ 2t)
dt i+d(3t−t3)
dt j
Step 2: Calculate the derivatives of the components of r(t).
Step 2: v(t) = (8t+ 2)i+ (3 −3t2)j
Step 3: Find the magnitude of the velocity at t= 2 s.
Step 3: |v(2)|=p(8(2) + 2)2+ (3 −3(2)2)2
Step 4: Simplify the expression and calculate the final answer.
Step 4: |v(2)|=p(16 + 2)2+ (3 −12)2=p182+ 92=√324 + 81 = √405 = 3√45 = 3√5 m/s
Therefore, the magnitude of the velocity of the particle at t= 2 s is 3√5
m/s.
32
Question 35
Question
A projectile is launched with an initial speed of 50 m/s at an angle of 30◦above
the horizontal. Find the time it takes for the projectile to reach its maximum
height.
Solution
Step 1: Break the initial velocity into its horizontal and vertical components.
Let’s denote the initial velocity of 50 m/s as v0and the angle of 30◦as θ. The
horizontal component of velocity (v0x) is given by:
v0x=v0cos(θ)
Plugging in the values, we get:
v0x= 50 cos(30◦)
v0x= 50 ×√3
2
v0x= 25√3 m/s
The vertical component of velocity (v0y) is given by:
v0y=v0sin(θ)
Plugging in the values, we get:
v0y= 50 sin(30◦)
v0y= 50 ×1
2
v0y= 25 m/s
Step 2: Calculate the time it takes for the projectile to reach its maximum
height. The time taken to reach the maximum height can be calculated using
the vertical component of velocity:
vf=v0y−g·t
At the maximum height, the vertical velocity is 0 m/s (vf= 0). Therefore, we
can solve for the time t:
0 = 25 −9.8·t
9.8t= 25
t=25
9.8
t≈2.55 s
Therefore, the time it takes for the projectile to reach its maximum height
is approximately 2.55 seconds.
33
Question 2
Question
A soccer player kicks a ball from the corner of a field. The ball follows a parabolic
path and lands 35 meters from the player. The initial velocity of the ball is 20
m/s at an angle of 30◦above the horizontal. How high above the ground is the
ball kicked?
(Note: Assume the acceleration due to gravity is 9.81 m/s2.)
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity v0can be resolved into its horizontal component v0xand
vertical component v0yas follows:
v0x=v0cos θ= 20 m/s ·cos 30◦= 17.32 m/s
v0y=v0sin θ= 20 m/s ·sin 30◦= 10.00 m/s
Step 2: Determine the time of flight. Since the time of flight for an ob-
ject launched at an angle in a uniform gravity field is twice the time to reach
maximum height, we can find the time of flight as follows:
tflight = 2 ×v0y
g= 2 ×10.00 m/s
9.81 m/s2= 2.04 s
Step 3: Determine the maximum height. The maximum height hmax reached
by the ball can be calculated using the formula:
hmax =v2
0y
2g=(10.00 m/s)2
2×9.81 m/s2= 5.10 m
Step 4: Calculate the distance from the player to the point where the ball
lands. The distance dfrom the player to where the ball lands is given by:
d=v0x×tflight = 17.32 m/s ×2.04 s = 35.28 m
Step 5: Determine the height above the ground where the ball is kicked.
Since the ball lands 35 meters away, the height above the ground where the ball
is kicked can be calculated as the vertical displacement of the ball at that point:
hkicked =hmax+1
2gt2
flight = 5.10 m+0.5×9.81 m/s2×(2.04 s)2= 5.10 m+20.16 m = 25.26 m
Therefore, the ball is kicked 25.26 meters above the ground.
2
Question 3
Question
A particle moves along a curved path in the xy plane. At a certain instant, the
particle’s position vector r is given by r = (3t2−2)ˆ
i+ (4t+ 1)ˆ
j, where tis in
seconds and the position coordinates are in meters. Determine the particle’s
velocity and acceleration at t= 2 s.
Solution
Step 1: To find the particle’s velocity, we differentiate the position vector r with
respect to time t:
v =dr
dt
Step 2: Differentiating the components of r:
d
dt[(3t2−2)ˆ
i]=6tˆ
i
d
dt[(4t+ 1)ˆ
j]=4ˆ
j
Therefore, the velocity vector at any time tis given by:
v = 6tˆ
i+ 4ˆ
j
Step 3: Now, to find the particle’s acceleration, we differentiate the velocity
vector v with respect to time t:
a =dv
dt
Step 4: Differentiating the components of v:
d
dt[6tˆ
i]=6ˆ
i
d
dt[4ˆ
j] = 0
Therefore, the acceleration vector at any time tis given by:
a = 6ˆ
i
Step 5: Finally, we evaluate the velocity and acceleration vectors at t= 2 s:
v(2) = 6(2)ˆ
i+ 4ˆ
j= 12ˆ
i+ 4ˆ
jm/s
a(2) = 6ˆ
im/s2
Therefore, at t= 2 s, the particle’s velocity is 12ˆ
i+ 4ˆ
jm/s and the acceler-
ation is 6ˆ
im/s2.
3
Question 4
Question
A golf ball is hit with an initial velocity of 30 m/s at an angle of 45 degrees
above the horizontal. Ignoring air resistance, calculate the maximum height the
ball will reach and the total time it takes to reach the ground.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity of the golf ball can be resolved into horizontal and vertical
components as follows: The initial horizontal component of velocity v0xis given
by: v0x=v0·cos θwhere v0= 30 m/s and θ= 45◦. So, v0x= 30 ·cos 45◦.
The initial vertical component of velocity v0yis given by: v0y=v0·sin θ
where v0= 30 m/s and θ= 45◦. So, v0y= 30 ·sin 45◦.
Step 2: Calculate the time taken to reach the maximum height. The time
taken to reach the maximum height can be calculated using the vertical com-
ponent of motion. The equation to determine the time taken to reach the
maximum height is:
v0y=vy−gt
Where v0yis the initial vertical component of velocity, vyis the vertical compo-
nent of velocity at the maximum height (which will be zero m/s), g= 9.81 m/s2
is the acceleration due to gravity, and tis the time taken to reach the maximum
height.
Step 3: Calculate the maximum height the golf ball will reach. The max-
imum height reached by the golf ball can be calculated using the equation for
vertical displacement:
y=v0y·t−1
2gt2
where yis the maximum height, v0yis the initial vertical component of velocity,
tis the time taken to reach the maximum height, and gis the acceleration due
to gravity.
Step 4: Calculate the total time taken to reach the ground. The total time
taken by the golf ball to reach the ground can be calculated using the formula
for the time of flight in projectile motion:
T=2v0sin θ
g
where Tis the total time of flight, v0is the initial velocity, θis the angle of
projection, and gis the acceleration due to gravity.
4
Question 5
Question
A soccer player kicks a ball from the ground with an initial velocity of 20 m/s
at an angle of 35◦above the horizontal. The ball lands on the ground 80 meters
away. Calculate the maximum height reached by the ball during its flight.
Solution
Step 1: Split the initial velocity into its horizontal and vertical components. The
horizontal component of the initial velocity can be calculated as v0x=v0cos(θ),
where v0= 20 m/s and θ= 35◦.
v0x= 20 cos(35◦)≈16.45 m/s
The vertical component of the initial velocity can be calculated as v0y=
v0sin(θ).
v0y= 20 sin(35◦)≈11.47 m/s
Step 2: Determine the time of flight. The time of flight tcan be determined
using the vertical component of the motion. The equation of motion in the
vertical direction is y=v0yt−1
2gt2, where yis the maximum height reached by
the ball and g= 9.81 m/s2is the acceleration due to gravity. At the maximum
height, the vertical component of the velocity is zero. Therefore, v0y−gt = 0.
t=v0y
g=11.47
9.81 ≈1.17 s
Step 3: Calculate the maximum height reached by the ball. Substitute the
time of flight into the equation for vertical motion.
y=v0yt−1
2gt2= 11.47 ×1.17 −1
2×9.81 ×(1.17)2≈6.49 m
Therefore, the maximum height reached by the ball during its flight is ap-
proximately 6.49 meters.
Question 6
Question
A baseball player hits a ball with an initial velocity of 30 m/s at an angle of
30 degrees above the horizontal. What is the maximum height the ball reaches
during its flight? (Assume the ball is hit from ground level and air resistance is
negligible.)
5
Solution
Step 1: Resolve the initial velocity into horizontal and vertical components. The
horizontal component v0xis given by: v0x=v0cos(θ) where v0= 30 m/s is the
initial velocity and θ= 30◦is the angle. Therefore, v0x= 30 ×cos(30◦).
Step 2: Calculate the horizontal component of the initial velocity. v0x=
30 ×cos(30◦) = 30 ×√3
2= 15√3 m/s.
Step 3: Use the kinematic equation to find the time taken to reach the
maximum height. The vertical component v0yis given by: v0y=v0sin(θ).
Substitute v0= 30 m/s and θ= 30◦to find v0y.
Step 4: Calculate the vertical component of the initial velocity. v0y= 30 ×
sin(30◦) = 30 ×1
2= 15 m/s.
Step 5: Use the equation vf=vi+at for vertical motion to find the time
taken to reach the maximum height. Since the ball reaches its maximum height,
the final velocity at this point is 0. Therefore, 0 = v0y−gt.
Step 6: Solve for the time t.t=v0y
g=15
9.8.
Step 7: Calculate the time t.t≈1.53 s.
Step 8: Use the equation y=v0yt−1
2gt2to find the maximum height.
y= 15 ×1.53 −1
2×9.8×(1.53)2.
Step 9: Calculate the maximum height y.y≈22.95 m.
Therefore, the maximum height the ball reaches during its flight is approxi-
mately 22.95 meters.
Question 7
Question
A baseball pitcher throws a fastball with an initial velocity of 30 m/s at an angle
of 30◦above the horizontal. At the instant the baseball leaves the pitcher’s hand,
a batter standing at home plate 18 m away sees the ball. Will the ball pass over
the plate or will it hit the ground first? Assume the baseball is in projectile
motion and air resistance is negligible.
Solution
Step 1: Break the initial velocity vector into its horizontal and vertical compo-
nents. The horizontal component of the velocity is given by:
v0x=v0cos(θ) = 30 m/s ·cos(30◦)≈25.98 m/s
The vertical component of the velocity is given by:
v0y=v0sin(θ) = 30 m/s ·sin(30◦)≈15 m/s
Step 2: Determine the time it takes for the ball to reach the batter. Let’s
consider the vertical motion of the baseball. The vertical position of the ball at
6
time tis given by:
y=v0yt−1
2gt2
where g= 9.81 m/s2is the acceleration due to gravity.
Since the ball is at ground level when it reaches the batter, y= 0 and we
can solve for t:
0 = 15t−1
2·9.81 ·t2
t(15 −4.905t)=0
t= 0 s or t≈3.06 s
Step 3: Determine the horizontal distance the ball travels in this time. The
horizontal distance the ball travels is given by:
x=v0xt
x= 25.98 m/s ·3.06 s ≈79.49 m
Since the batter is only 18 m away, the ball will hit the ground before passing
over the plate.
Question 8
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. Find the maximum height above the ground
reached by the projectile. Take g= 9.8 m/s2.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity can be resolved into two components: v0x=v0cos(θ) and
v0y=v0sin(θ), where v0= 20 m/s and θ= 30◦. Thus, v0x= 20 m/s ·cos(30◦)
and v0y= 20 m/s ·sin(30◦).
Step 2: Find the time taken to reach the maximum height. Using vy=
v0y−gt and vy= 0 at the maximum height, we have 0 = 20 sin(30◦)−9.8t.
Solving for t, we get t=20 sin(30◦)
9.8.
Step 3: Calculate the maximum height. The maximum height hcan be
found using the formula h=ymax =v0yt−1
2gt2. Substitute the values of v0y
and tto find h. Remember that at the highest point vy= 0, so v0y−gt = 0.
Thus, h= 20 sin(30◦)·20 sin(30◦)
9.8−1
2·9.820 sin(30◦)
9.82. Simplify this expression
to find the maximum height.
7
Question 9
Question
A baseball is hit at an angle of 30◦above the horizontal with an initial speed of
40 m/s. At the same time, a gust of wind starts blowing the baseball horizontally
at a constant speed of 5 m/s. Calculate the maximum height above its starting
point that the baseball reaches.
Solution
Step 1: Resolve the initial velocity of the baseball into its horizontal and vertical
components. Let v0xbe the initial velocity in the x-direction and v0ybe the
initial velocity in the y-direction, then: v0x=v0cos(30◦) = 40 cos(30◦)≈34.64
m/s v0y=v0sin(30◦) = 40 sin(30◦)≈20 m/s
Step 2: Determine the time taken for the baseball to reach its maximum
height. The time taken to reach the maximum height can be calculated using
the equation vfy =v0y−gt, where vf y is the final velocity in the y-direction
(which is 0 m/s at the maximum height), gis the acceleration due to gravity
(approximately -9.81 m/s2), and tis the time.
0 = 20 −9.81t
t=20
9.81 ≈2.04 seconds
Step 3: Calculate the maximum height reached by the baseball. The maxi-
mum height can be determined by using the equation y=v0yt−1
2gt2, where y
is the height, v0yis the initial velocity in the y-direction, tis the time calculated
in step 2, and gis the acceleration due to gravity.
y= 20(2.04) −1
2(9.81)(2.04)2
y≈20.4 meters
Therefore, the baseball reaches a maximum height of approximately 20.4
meters above its starting point.
Question 10
Question
A baseball is hit with an initial velocity of 30 m/s at an angle of 45 degrees
above the horizontal. Find the maximum height the baseball reaches and the
total time it is in the air before hitting the ground. Assume air resistance is
negligible and take the acceleration due to gravity as 9.81 m/s2.
8
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity of the baseball can be resolved into its horizontal (v0x) and
vertical (v0y) components as follows:
v0x=v0cos θ= 30 cos 45◦≈21.21 m/s
v0y=v0sin θ= 30 sin 45◦≈21.21 m/s
Step 2: Find the time to reach maximum height. The time to reach the
maximum height (tpeak) can be calculated using the vertical component of the
initial velocity and the acceleration due to gravity:
vpeak = 0
vpeak =v0y−gtpeak
0 = 21.21 −9.81tpeak
tpeak =21.21
9.81 ≈2.16 s
Step 3: Calculate the maximum height. The maximum height (ymax) can
be found using the equation for vertical position:
ymax =v0ytpeak −1
2gt2
peak
ymax = 21.21 ×2.16 −1
2×9.81 ×(2.16)2
ymax ≈23.15 m
Step 4: Determine the total time in the air. The total time the baseball is
in the air can be calculated as twice the time to reach the peak:
Total time = 2tpeak = 2 ×2.16
Total time ≈4.32 s
Therefore, the baseball reaches a maximum height of approximately 23.15
m and is in the air for approximately 4.32 seconds before hitting the ground.
Question 11
Question
A ball is thrown from the edge of a cliff at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. The cliff is 50 m high. How far from the base
of the cliff will the ball hit the ground?
9
Solution
Let’s break down the motion of the ball into its horizontal and vertical compo-
nents. We can then solve for the time it takes for the ball to hit the ground and
use that time to find the horizontal distance traveled.
Step 1: Find the time it takes for the ball to hit the ground.
We know the initial vertical velocity of the ball is 20 sin(30◦) m/s.
The acceleration due to gravity is −9.8 m/s2.
The ball will hit the ground when its vertical position is 0.
Using the equation of motion:
yf=yi+viyt+1
2at2
we can set yf= 0, yi= 50 m, viy= 20 sin(30◦) m/s, and a=−9.8 m/s2to
solve for t.
0 = 50 + 20 sin(30◦)t−4.9t2
Solving this quadratic equation gives t≈3.25 s.
Step 2: Find the horizontal distance traveled by the ball.
The initial horizontal velocity of the ball is 20 cos(30◦) m/s.
The horizontal distance the ball travels is given by:
d=vixt
Substitute vix= 20 cos(30◦) m/s and t≈3.25 s to find:
d≈(20 cos(30◦)·3.25) m
Calculating this value, we find that the ball will hit the ground approximately
56.4 m from the base of the cliff.
Question 12
Question
A soccer player kicks a ball with an initial velocity of 20 m/s at an angle of 30
degrees above the horizontal. How far does the ball travel horizontally before it
hits the ground? (Assume no air resistance and g= 9.8 m/s2)
10
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity can be broken down into its horizontal and vertical com-
ponents using trigonometric functions. The horizontal component is given by
v0x=v0cos(θ) and the vertical component is v0y=v0sin(θ), where v0= 20 m/s
and θ= 30◦.
Step 2: Calculate the time of flight. The time taken for the ball to reach the
ground can be determined using the vertical component of the velocity. The
time of flight tcan be found using the equation y=v0yt+1
2gt2, where y= 0
(since the ball starts and ends at ground level).
Step 3: Calculate the horizontal distance traveled. The horizontal distance
traveled by the ball can be calculated using the horizontal component of the
velocity and the time of flight. The horizontal distance xis given by x=v0xt.
Step 4: Substitute the values and solve for the horizontal distance. Substi-
tute the given values (v0= 20 m/s, θ= 30◦,g= 9.8 m/s2) into the equations
derived in Steps 1 to 3, and solve for the horizontal distance the ball travels
before hitting the ground.
Question 13
Question
A particle moves in the xy plane according to the following set of equations:
x= (3.00 m/s)tcos(30◦)
y= (3.00 m/s)tsin(30◦)−1
2(9.80 m/s2)t2
At what time does the particle cross the xaxis for the first time?
Solution
Step 1: Find the xcoordinate of the particle crossing the xaxis for the first
time. The particle crosses the xaxis when y= 0, so we substitute y= 0 into
the given equations:
0 = (3.00 m/s)tsin(30◦)−1
2(9.80 m/s2)t2
0 = (3.00 m/s)t1
2−1
2(9.80 m/s2)t2
0 = 1.50t−4.90t2
Step 2: Solve the quadratic equation 1.50t−4.90t2= 0. Factoring out t, we
get:
t(1.50 −4.90t)=0
11
Thus, t= 0 or t=1.50
4.90 ≈0.31 s.
Step 3: Since the particle is already moving in the positive xdirection at
t= 0, we can conclude that the particle crosses the xaxis for the first time at
t≈0.31 s.
Question 14
Question
A soccer player kicks a ball from the ground with an initial velocity of 20 m/s
at an angle of 30 degrees above the horizontal. The ball lands on the ground
60 meters away. Calculate: a) The time of flight of the ball. b) The maximum
height the ball reaches. c) The magnitude of the ball’s velocity just before it
hits the ground.
Solution
a) To find the time of flight, we can use the equation for the horizontal displace-
ment of a projectile:
∆x=Vix·t
where ∆xis the horizontal displacement, Vixis the initial horizontal velocity,
and tis the time of flight.
Step 1: Calculate the initial horizontal velocity component:
Vix=Vi·cos θ
Vix= 20 m/s ·cos(30◦)
Vix= 20 m/s ·√3
2
Vix= 10√3 m/s
Step 2: Substitute the known values into the equation and solve for t:
60 = 10√3 m/s ·t
t=60
10√3
t=6√3
3
t= 2√3 seconds
Therefore, the time of flight of the ball is 2√3 seconds.
12
b) To find the maximum height, we can use the vertical component of the
initial velocity and the time of flight along with the equation for vertical dis-
placement of a projectile:
∆y=Viy·t+1
2·a·t2
where: - ∆yis the vertical displacement, - Viyis the initial vertical velocity
component, - tis the time of flight, and - ais the acceleration due to gravity.
Step 3: Calculate the initial vertical velocity component:
Viy=Vi·sin θ
Viy= 20 m/s ·sin(30◦)
Viy= 20 m/s ·1
2
Viy= 10 m/s
Step 4: Substitute the known values into the equation and solve for the
maximum height:
∆y= 10 m/s ·2√3 + 1
2·(−9.8 m/s2)·(2√3)2
∆y= 20√3−29.4×3
∆y= 20√3−88.2
∆y≈3.188 meters
Therefore, the maximum height the ball reaches is approximately 3.188 me-
ters.
c) The magnitude of the ball’s velocity just before it hits the ground is equal
to the initial speed since no horizontal force acts on the ball.
Hence, the magnitude of the ball’s velocity just before it hits the ground is
20 m/s.
Question 15
Question
A projectile is launched from the ground at an angle of 45◦above the horizontal
with an initial speed of 20 m/s. Find the total time the projectile is in the
air and the maximum height it reaches. Your answer should include both the
mathematical expressions and the numerical values.
13
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity v0can be resolved into horizontal and vertical components
as follows:
v0x=v0cos(θ) and v0y=v0sin(θ)
where v0xis the initial velocity in the x-direction and v0yis the initial velocity
in the y-direction. Here, v0= 20 m/s and θ= 45◦.
Step 2: Calculate the initial velocity components v0xand v0y. Substitute
v0= 20 m/s and θ= 45◦into the equations:
v0x= 20 cos(45◦) = 20 ×√2
2= 10√2 m/s
v0y= 20 sin(45◦) = 20 ×√2
2= 10√2 m/s
Step 3: Find the time of flight. The time of flight can be determined by
using the y-component of the displacement equation:
y=v0yt−1
2gt2
Since the projectile returns to the same height it was launched from, y= 0 and
solving for tgives:
0 = 10√2t−1
2(9.8)t2
Solving for twe get two solutions: t= 0 or t≈2.03 s. Since t= 0 is the initial
time, the total time the projectile is in the air is approximately 2.03 seconds.
Step 4: Find the maximum height. The maximum height can be determined
by using the y-component of the displacement equation:
y=v0yt−1
2gt2
at the top of its trajectory, the vertical component of velocity is 0. Therefore,
v0y−gt = 0.
t=v0y
g=10√2
9.8≈1.43 s
Substitute t= 1.43 s into the equation gives:
y= 10√2×1.43 −1
2×9.8×(1.43)2≈14.1 m
Thus, the total time the projectile is in the air is approximately 2.03 seconds
and the maximum height it reaches is approximately 14.1 meters.
14
Question 16
Question
A baseball is hit with an initial velocity of 30 m/s at an angle of 40◦above the
horizontal. Calculate the maximum height the baseball reaches and the total
horizontal distance it travels before hitting the ground. Assume air resistance
is negligible and the acceleration due to gravity is 9.81 m/s2.
Solution
Step 1: Break the initial velocity into its horizontal and vertical components.
The horizontal component of velocity, v0x, can be calculated as:
v0x=v0cos(θ)
v0x= 30 cos(40◦)
v0x≈23.05 m/s
The vertical component of velocity, v0y, can be calculated as:
v0y=v0sin(θ)
v0y= 30 sin(40◦)
v0y≈19.34 m/s
Step 2: Calculate the time taken to reach maximum height. The time taken
to reach maximum height can be calculated using the vertical component of
velocity and acceleration due to gravity:
vfy =v0y−gt
0 = 19.34 −9.81t
t=19.34
9.81
t≈1.97 s
Step 3: Calculate the maximum height reached by the baseball. The max-
imum height can be calculated using the vertical component of velocity and
time:
ymax =v0yt−1
2gt2
ymax = 19.34 ·1.97 −1
2·9.81 ·(1.97)2
ymax ≈19.08 m
15
Step 4: Calculate the total horizontal distance traveled by the baseball. The
total horizontal distance can be calculated using the horizontal component of
velocity and time:
xtotal =v0x·t
xtotal = 23.05 ·1.97
xtotal ≈45.34 m
Therefore, the baseball reaches a maximum height of approximately 19.08 m
and travels a total horizontal distance of approximately 45.34 m.
Question 17
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. Find the total time of flight, the maximum
height reached, and the horizontal range of the projectile. Assume the acceler-
ation due to gravity is −9.81 m/s2.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity of the projectile can be resolved into horizontal (v0x) and
vertical (v0y) components as:
v0x=v0cos(θ)
v0y=v0sin(θ)
Given that v0= 20 m/s and θ= 30◦, we can calculate v0xand v0yas follows:
v0x= 20 ×cos(30◦) = 20 ×√3
2= 10√3 m/s
v0y= 20 ×sin(30◦) = 20 ×1
2= 10 m/s
Step 2: Find the time of flight. The total time of flight (T) can be found
using the y-component of the motion. At the maximum height, the vertical
component of the velocity is zero. The time taken to reach the maximum height
is equal to the time taken for the projectile to descend from this point to the
ground.
Using the equation vy=v0y−gt, where vy= 0 at the maximum height, we
can find the time taken to reach the maximum height:
0 = 10 −9.81t
t=10
9.81 = 1.02 s
16
Therefore, the total time of flight is twice the time taken to reach the max-
imum height:
T= 2t= 2 ×1.02 = 2.04 s
Step 3: Find the maximum height reached. To find the maximum height
(hmax) reached by the projectile, we use the equation:
hmax =v2
0y/(2g)
Substitute the known values to find hmax:
hmax = (10)2/(2 ×9.81) = 5.1 m
Step 4: Find the horizontal range. The horizontal range of the projectile is
the horizontal component of the velocity multiplied by the total time of flight:
R=v0x·T= 10√3·2.04 ≈20.8 m
Therefore, the total time of flight is 2.04 s, the maximum height reached is
5.1 m, and the horizontal range of the projectile is approximately 20.8 m.
Question 18
Question
A car travels along a straight road with an initial velocity of 20 m/s. It acceler-
ates at a constant rate of 2 m/s2for 5 seconds, and then continues at a constant
velocity for another 10 seconds. At this point, the driver sees an obstacle ahead
and applies the brakes, causing the car to decelerate at a rate of 3 m/s2. Find
the total distance the car travels during the entire trip.
Solution
Step 1: Find the distance traveled during acceleration phase. The distance d1
traveled during the acceleration phase can be found using the equation d1=
vit+1
2at2, where viis the initial velocity, ais the acceleration, and tis the
time. Plugging in the values, we get: d1= (20 m/s)(5 s) + 1
2(2 m/s2)(5 s)2
d1= 100 m + 25 m d1= 125 m
Step 2: Find the distance traveled during constant velocity phase. The dis-
tance d2traveled during the constant velocity phase can be found by multiplying
the velocity by time. d2= (20 m/s)(10 s) d2= 200 m
Step 3: Find the distance traveled during deceleration phase. The distance
d3traveled during the deceleration phase can be found using the equation d3=
vft−1
2at2, where vfis the final velocity, ais the deceleration, and tis the
time. First, calculate the final velocity before braking: Vf=Vi+at = 20 m/s +
2 m/s2·5 s = 30 m/s.Then, calculate the distance traveled during deceleration:
d3= (30 m/s)(10 s) −1
2(3 m/s2)(10 s)2d3= 300 m −150 m d3= 150 m
17
Step 4: Find the total distance traveled. The total distance traveled is the
sum of the distances traveled during each phase: Total distance = d1+d2+d3=
125 m + 200 m + 150 m = 475 m
Therefore, the total distance the car travels during the entire trip is 475
meters.
Question 19
Question
A car travels along a straight road and then around a curve that is part of a
circular racetrack. The car covers the second half of the curve in half the time
it covers the first half. If the total distance covered is 1.5 km and the average
speed is 60 km/h, what is the radius of the curve?
Solution
Step 1: Let’s denote the distance covered in the first half of the curve as d1and
the distance covered in the second half as d2. Since the car’s average speed is
60 km/h, we can write:
d1+d2= 1.5 km
The time taken to cover the first half, t1, is twice the time taken to cover the
second half, t2. We can write:
t1= 2t2
Step 2: The average speed of the car is given by:
Average speed = total distance
total time =d1+d2
t1+t2
Step 3: We can rewrite the average speed using the relationship between
distance, time, and speed:
Average speed = d1+d2
t1+t2
=d1+d2
d1
v+d2
2v
where vis the speed of the car.
Step 4: Substituting the known values into the equation:
60 = 1.5
d1
60 +d1
120
Step 5: Solving for d1:
60 = 1.5
3
2d1
d1= 1 km
18
Step 6: Substituting d1back into the total distance equation:
1 + d2= 1.5
d2= 0.5 km
Step 7: The time taken to cover the first half is:
t1=1
60 =1
60 hour
Step 8: The time taken to cover the second half is:
t2=0.5
60 =1
120 hour
Step 9: Using the formula for centripetal acceleration, a=v2
r, where vis
the speed of the car and ris the radius of the curve, we have:
a=(60 km/h)2
r
Step 10: At the midway point of the curve, the car’s acceleration must be
equal to the acceleration due to gravity to keep the car moving along the curve.
Thus,
(60 km/h)2
r=g
where g= 9.81 m/s2is the acceleration due to gravity.
Step 11: Converting the speed to m/s:
60 km/h = 60 ×1000
3600 m/s = 500
3m/s
Step 12: Substituting the values into the equation:
(500
3)2
r= 9.81
r=(500
3)2
9.81
r≈278.3 m
Therefore, the radius of the curve is approximately 278.3 m.
Question 20
Question
A baseball player hits a ball that is caught by an outfielder 3.5 seconds later. If
the ball was hit at an angle of 30◦above the horizontal and was caught at the
same height it was hit, what was the initial velocity of the ball?
19
Solution
Step 1: Identify known variables and convert angle to radians. Step 2: Break
initial velocity into horizontal and vertical components. Step 3: Use the vertical
motion equation to find the time the ball spends in the air. Step 4: Use the
horizontal motion equation to find the initial velocity. Step 5: Calculate the
initial velocity of the ball.
Step 1: The known variables are time t= 3.5 s, angle θ= 30◦, and accel-
eration due to gravity g= 9.8 m/s2. Converting the angle to radians, we have
θ= 30◦×π
180 =π
6radians.
Step 2: The initial velocity v0can be broken down into horizontal v0xand
vertical v0ycomponents:
v0x=v0cos(θ) and v0y=v0sin(θ)
Step 3: In the vertical direction, the motion equation is:
y=v0yt−1
2gt2
Since the ball was caught at the same height it was hit, the height is 0, and the
equation simplifies to:
0 = v0sin(θ)t−1
2gt2
Solving for t, we get:
t=2v0sin(θ)
g
Step 4: In the horizontal direction, the motion equation is:
x=v0xt
Substitute v0x=v0cos(θ) and t=2v0sin(θ)
g:
x=v0cos(θ)·2v0sin(θ)
g
Step 5: The initial velocity of the ball can be found by solving for v0:
x=2v2
0sin(θ) cos(θ)
g
v2
0=gx
2 sin(θ) cos(θ)
v0=rgx
2 sin(θ) cos(θ)
Substitute x= 0 (initial position) and θ=π
6:
v0=s9.8×0
2 sin π
6cos π
6= 0
Therefore, the initial velocity of the ball was 0 m/s.
20
Question 21
Question
A tennis ball is hit at an angle of 30◦above the horizontal from a height of 2
meters at an initial speed of 20 m/s. How far from the base of the wall does the
ball land if the wall is 10 meters high?
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical compo-
nents. The initial velocity can be broken down into its horizontal and verti-
cal components as follows: Vix=Vicos θ= 20 m/s ×cos(30◦) = 17.32 m/s
Viy=Visin θ= 20 m/s ×sin(30◦) = 10 m/s
Step 2: Determine the time taken to reach the top of the wall. We can
calculate the time taken to reach the highest point using the vertical component
of the initial velocity. The equation we can use is: Vf=Vi+at Where Vfis the
final vertical velocity (0 m/s at the top), ais the acceleration due to gravity (-9.8
m/s2), and tis the time taken. Substitute the known values into the equation
and solve for t: 0 = 10 −9.8t t =10
9.8= 1.02 s
Step 3: Calculate the horizontal distance the ball travels in the air. The
horizontal distance the ball travels is given by the equation: d=Vix×tSubsti-
tute the values of Vixand tinto the equation: d= 17.32 m/s ×1.02 s = 17.66
m.
Step 4: Determine the height reached by the ball at the point it lands.
The height reached by the ball when it lands can be calculated using the ver-
tical component of the initial velocity and the time taken to hit the wall. The
equation we can use is: H=Viyt−1
2gt2Substitute the known values into the
equation and solve for H:H= 10 ×1.02 −1
2×9.8×(1.02)2= 5.1−5 = 0.1 m
Step 5: Determine the distance from the base of the wall where the ball lands.
The horizontal distance from the wall where the ball lands is the sum of the dis-
tance traveled horizontally and the distance it fell vertically. Given that the wall
is 10 meters high, the total distance is: 17.66 m+p(10 m −2 m)2+ (0.1 m)2=
17.66 m + √64.1≈25.33 m
Therefore, the ball lands approximately 25.33 meters from the base of the
wall.
Question 22
Question
A projectile is launched at an angle of 30◦above the horizontal with an initial
speed of 20 m/s. Calculate the time it takes for the projectile to reach its
maximum height. Assume that air resistance is negligible and take g= 9.8
m/s2.
21
Solution
Step 1: Resolve the initial velocity into its xand ycomponents. The initial
velocity is given as 20 m/s at an angle of 30◦above the horizontal. Resolving
this into its xand ycomponents, we get: V0x= 20 cos 30◦= 20 ·√3
2= 10√3
m/s, V0y= 20 sin 30◦= 20 ·1
2= 10 m/s.
Step 2: Calculate the time it takes to reach maximum height. At the highest
point, the vertical component of velocity becomes 0. We can use the equation:
Vy=V0y−gt, 0 = 10 −9.8t,t=10
9.8≈1.02 seconds.
Therefore, it takes approximately 1.02 seconds for the projectile to reach its
maximum height.
Question 23
Question
A football is kicked with an initial velocity of 20 m/s at an angle of 30◦above
the horizontal. Find the maximum height the football reaches and the time it
takes to reach this height.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
Let v0= 20 m/s be the initial velocity and θ= 30◦be the angle above the
horizontal.
The initial horizontal velocity v0xis given by v0x=v0cos θ.
The initial vertical velocity v0yis given by v0y=v0sin θ.
Step 2: Determine the time it takes to reach the maximum height.
The time it takes to reach the maximum height is given by the vertical
motion equation y=v0yt−1
2gt2, where y= 0 (maximum height) and
g= 9.81 m/s2is the acceleration due to gravity.
Substitute v0yand ginto the equation and solve for t.
Step 3: Find the maximum height.
The maximum height is given by the equation ymax =v0yt−1
2gt2using
the time tobtained in step 2.
Substitute v0y,t, and ginto the equation and calculate ymax.
22
Question 24
Question
A particle moves in the xy-plane with an acceleration given by a = 4tˆ
i+ 3ˆ
j
m/s2, where ˆ
iand ˆ
jare unit vectors in the xand ydirections, respectively. At
t= 0, the particle is at the origin with a velocity of 5ˆ
im/s. Find the speed of
the particle when it reaches a position 10 m along the x-axis at an angle of 30◦
above it.
Solution
Step 1: The components of velocity are obtained by integrating the components
of acceleration with respect to time.
Zaxdt =Z4t dt = 2t2+C1
Zaydt =Z3dt = 3t+C2
Given that at t= 0, the particle has a velocity of 5ˆ
im/s, we have:
vx(0) = 5ˆ
i= 0ˆ
i+C1⇒C1= 5ˆ
i
vy(0) = 0ˆ
j+C2= 0ˆ
j⇒C2= 0ˆ
j
Thus, the velocity components are:
vx= 2t2+ 5
vy= 3t
Step 2: We can find speed from the velocity components.
v=qv2
x+v2
y
v(t) = p(2t2+ 5)2+ (3t)2
At the final position, x= 10 m and y= 10 tan 30◦= 5√3 m. We need to find
the time tat which x= 10 m by solving x=Rvxdt =R(2t2+ 5) dt.
10 = Z(2t2+ 5) dt =2
3t3+ 5t+C3
Given initial velocity vx(0) = 5, we can evaluate the constant C3:
5=03+ 5 ∗0 + C3⇒C3= 5
Thus, x(t) = 2
3t3+ 5t+ 5.
23
Step 3: To find the time tat which x= 10 m, we solve for tin the equation
x(t) = 10.
10 = 2
3t3+ 5t+ 5
2
3t3+ 5t−5 = 0
This cubic equation must be solved numerically. Once tis found and substituted
back into the expression for v(t), the speed of the particle when it reaches the
specified position can be calculated.
Question 25
Question
A football is kicked at an angle of 30◦above the horizontal with an initial
speed of 20 m/s. How far does the football travel horizontally before hitting the
ground? (Assume no air resistance)
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity v0can be resolved into two components: v0x(horizontal
component) and v0y(vertical component).
v0x=v0cos(θ) = (20 m/s) cos(30◦) = 17.32 m/s
v0y=v0sin(θ) = (20 m/s) sin(30◦) = 10 m/s
Step 2: Determine the time of flight. The time taken for the football to
reach the ground can be determined using the vertical component of motion.
We will use vyf =v0y+at and y=v0yt+1
2at2. When the football hits the
ground, y= 0 and vyf = 0.
0 = 10 m/s + (−9.8 m/s2)t
t=10 m/s
9.8 m/s2= 1.02 s
Step 3: Calculate the horizontal distance traveled. The horizontal distance
xtraveled by the football can be calculated using the formula x=v0xt.
x= (17.32 m/s)(1.02 s) = 17.66 m
Therefore, the football travels approximately 17.66 meters horizontally be-
fore hitting the ground.
24
Question 26
Question
A projectile is launched from the ground with an initial speed of 30 m/s at an
angle of 30◦above the horizontal. Find the maximum height reached by the
projectile.
Solution
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The horizontal component is v0x=v0cos(θ) and the vertical component is
v0y=v0sin(θ). Step 2: Calculate the time it takes for the projectile to reach
maximum height using the equation vf=vi+a∆t, where vf= 0 at maximum
height. Step 3: Use the time found in Step 2 to calculate the maximum height
using the formula ymax =yi+v0yt−1
2gt2, where yi= 0 and g= 9.81 m/s2is
the acceleration due to gravity.
Solution
Step 1: The initial velocity can be resolved into its horizontal and vertical
components:
v0x=v0cos(θ) = 30 m/s ×cos(30◦) = 30√3/2 m/s
v0y=v0sin(θ) = 30 m/s ×sin(30◦) = 15 m/s
Step 2: The time to reach maximum height can be found using the vertical
component of velocity:
vf=vi+a∆t
0 = 15 m/s −9.81 m/s2×t
t=15 m/s
9.81 m/s2
t≈1.53 s
Step 3: The maximum height can be found using the vertical motion equa-
tion:
ymax =yi+v0yt−1
2gt2
ymax = 0 + 15 m/s ×1.53 s −1
2×9.81 m/s2×(1.53 s)2
ymax ≈11.5 m
Therefore, the maximum height reached by the projectile is approximately
11.5 meters.
25
Question 27
Question
A particle moves along a curved path in the xy plane. The position of the
particle as a function of time is given by the equations:
x(t)=3t2−4t
y(t)=2t3−6t2
Find the velocity and acceleration vectors of the particle at time t= 2.
Solution
Step 1: Find the velocity vector by taking the derivative of the position vector
with respect to time.
Velocity, v(t) = dr
dt =d
dt(x(t)ˆ
i+y(t)ˆ
j)
v(t) = d
dt(3t2−4t)ˆ
i+d
dt(2t3−6t2)ˆ
j
v(t) = (6t−4) ˆ
i+ (6t2−12t)ˆ
j
Step 2: Find the acceleration vector by taking the derivative of the velocity
vector with respect to time.
Acceleration, a(t) = dv
dt =d
dt((6t−4) ˆ
i+ (6t2−12t)ˆ
j)
a(t) = (6)ˆ
i+ (12t−12)ˆ
j
Step 3: Evaluate the velocity and acceleration vectors at t= 2.
At t= 2 :
v(2) = (6(2) −4) ˆ
i+ (6(2)2−12(2)) ˆ
j= 8 ˆ
i+ 12 ˆ
j
a(2) = (6) ˆ
i+ (12(2) −12) ˆ
j= 6 ˆ
i+ 12 ˆ
j
Therefore, at t= 2, the velocity vector is 8 ˆ
i+ 12 ˆ
jand the acceleration
vector is 6 ˆ
i+ 12 ˆ
j.
Question 28
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 40 m/s. Determine: (a) the maximum height above the
ground reached by the projectile, (b) the total time it takes for the projectile
to return to the ground.
26
Solution
Let’s solve the problem step by step:
Step 1: Resolve the initial velocity into its horizontal and vertical
components. The initial velocity of the projectile can be resolved into hori-
zontal and vertical components as follows: The initial velocity in the x-direction
(v0x) is given by:
v0x=v0cos(θ)
where v0= 40 m/s is the initial speed and θ= 30◦is the launch angle. Substitute
v0= 40 m/s and θ= 30◦into the equation:
v0x= 40 m/s ·cos(30◦) = 34.64 m/s
The initial velocity in the y-direction (v0y) is given by:
v0y=v0sin(θ)
Substitute v0= 40 m/s and θ= 30◦into the equation:
v0y= 40 m/s ·sin(30◦) = 20 m/s
Step 2: Calculate the time to reach maximum height (tmax). The
time taken to reach maximum height can be found using the equation:
vmax =v0y−g·tmax
where vmax = 0 m/s at the maximum height and g= 9.8 m/s2is the acceleration
due to gravity. Substitute vmax = 0 m/s, v0y= 20 m/s, and g= 9.8 m/s2into
the equation:
0 = 20 m/s −9.8 m/s2·tmax
Solving for tmax:
tmax =20 m/s
9.8 m/s2= 2.04 s
Step 3: Calculate the maximum height above the ground. The
maximum height reached by the projectile can be found using the equation:
ymax =v0y·tmax −1
2·g·(tmax)2
Substitute v0y= 20 m/s, tmax = 2.04 s, and g= 9.8 m/s2into the equation:
ymax = 20 m/s ·2.04 s −1
2·9.8 m/s2·(2.04 s)2
Calculating ymax:
ymax = 20 m/s ·2.04 s −1
2·9.8 m/s2·(2.04 s)2= 20.4 m
Step 4: Calculate the total time of flight. The total time of flight can
be determined by finding the time it takes for the projectile to return to the
ground, which is twice
27
Question 29
Question
A particle moves in the xy plane such that its position vector is given by r =
(4t2−2t)i+ (2t3+t)j, where tis in seconds. Determine the velocity and
acceleration vectors of the particle when t= 2 s.
Solution
Step 1: To find the velocity vector, we need to differentiate the position vector
r with respect to time t.
v =dr
dt
Step 2: Differentiating each component of the position vector r:
v =d
dt[(4t2−2t)i+ (2t3+t)j]
v = (8t−2)i+ (6t2+ 1)j
Step 3: Now, we need to find the acceleration vector by differentiating the
velocity vector v with respect to time t.
a =dv
dt
Step 4: Differentiating each component of the velocity vector v:
a =d
dt[(8t−2)i+ (6t2+ 1)j]
a = 8i+ 12tj
Step 5: Now, substitute t= 2 s into the velocity and acceleration vectors to
find their values at t= 2 s.
vt=2 = (8(2) −2)i+ (6(2)2+ 1)j
vt=2 = 14i+ 25j
at=2 = 8i+ 12(2)j
at=2 = 8i+ 24j
Therefore, the velocity vector at t= 2 s is vt=2 = 14i+ 25jm/s, and the
acceleration vector at t= 2 s is at=2 = 8i+ 24jm/s
²
.
28
Question 30
Question
A particle moves in the xy plane such that its position vector at time tis given
by r(t) = (3t2+ 2t)ˆ
i−(4t+ 1)ˆ
j, where ˆ
iand ˆ
jare unit vectors in the xand
ydirections, respectively. Find the magnitude of the particle’s acceleration at
t= 2 s.
Solution
Step 1: Find the particle’s velocity at t= 2 s by differentiating the position
vector:
v(t) = dr
dt = (6t+ 2)ˆ
i−4ˆ
j
Substitute t= 2 s to find the velocity at t= 2 s:
v(2) = (6(2) + 2)ˆ
i−4ˆ
j= 14ˆ
i−4ˆ
j
Step 2: Find the particle’s acceleration at t= 2 s by differentiating the
velocity vector:
a(t) = dv
dt = 6ˆ
i
Therefore, the magnitude of the particle’s acceleration at t= 2 s is:
|a(2)|=|6ˆ
i|= 6 m/s2
Question 31
Question
A particle is initially at the origin at t= 0. The particle then moves in the
xy-plane and has a position vector r= (4t3−2t)i+ (6t2+ 4)j, where iand jare
unit vectors in the xand ydirections, respectively. Find the magnitude of the
particle’s acceleration vector at t= 2 seconds.
Solution
Step 1: Find the velocity vector v(t) by taking the derivative of the position
vector r(t) with respect to time:
v(t) =
dt = ddt[(4t3−2t)i+(6t2+4)j]
v(t) = (12t2−2)i+ (12t)j
29
Step 2: Find the acceleration vector a(t) by taking the derivative of the
velocity vector v(t) with respect to time:
a(t) =
dt = ddt[(12t2−2)i+(12t)j]
a(t) = (24t)i+ 12j
Step 3: Find the acceleration vector aat t= 2 seconds:
a= (24(2))i+ 12j
a= 48i+ 12j
Step 4: Calculate the magnitude of the acceleration vector:
|a|=p(48)2+ (12)2
|a|=√2304 + 144
|a|=√2448
|a| ≈ √2400
|a| ≈ 49.38 m/s2
Therefore, the magnitude of the particle’s acceleration vector at t= 2 sec-
onds is approximately 49.38 m/s2.
Question 32
Question
A projectile is launched with an initial speed of 30 m/s at an angle of 30◦above
the horizontal from the ground. Determine the time taken for the projectile to
reach the maximum height. (Use g= 9.81 m/s2for acceleration due to gravity)
Solution
Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial velocity (vi) can be resolved into horizontal (vix) and vertical (viy)
components using trigonometry:
vix =vi·cos(θ)
viy =vi·sin(θ)
where vi= 30 m/s and θ= 30◦.
Step 2: Calculate the time taken to reach the maximum height. At the peak
of the projectile’s trajectory, the vertical velocity component is zero. We can
30
use this information to find the time taken to reach the maximum height. The
vertical velocity at any time tis given by:
vfy =viy −gt
At the maximum height, vfy = 0:
0 = viy −gt
t=viy
g
Substitute viy and gto find the time taken to reach the maximum height.
Step 3: Substitute the known values and calculate the time. Calculate vix,
viy, and substitute them into the equation for time:
vix = 30 ·cos(30◦) = 25.98 m/s
viy = 30 ·sin(30◦) = 15 m/s
Now substitute viy into the equation:
t=15
9.81 = 1.53 seconds
Therefore, the time taken for the projectile to reach the maximum height is
1.53 seconds.
Question 33
Question
A projectile is launched from the ground at an angle of 30◦above the horizontal
with an initial speed of 20 m/s. At the highest point of its trajectory, the
projectile explodes into two fragments of equal mass. One fragment immediately
drops vertically, and the other fragment continues to move along the original
path after the explosion. What is the speed of the fragment that continues to
move along the original path just after the explosion?
Solution
Step 1: Resolve initial velocity into horizontal and vertical components: The
initial velocity of the projectile can be resolved into its horizontal and vertical
components as: Horizontal component: V0x=V0·cos(30◦) Vertical component:
V0y=V0·sin(30◦)
Given that V0= 20 m/s, we substitute to find: V0x= 20·cos(30◦) = 20·√3
2=
10√3 m/s V0y= 20 ·sin(30◦) = 20 ·1
2= 10 m/s
Step 2: Time to reach highest point: The time taken for the projectile to
reach the highest point can be found using the vertical component of velocity.
31
At the highest point, the vertical component of velocity becomes 0. We can use
this information to find the time taken. The vertical component of velocity can
be given by: Vy=V0y−gt where Vy= 0 at the highest point.
Substitute V0y= 10 m/s and g= 9.81 m/s2to solve for t: 0 = 10 −9.81t
t=10
9.81 ≈1.02 s
Step 3: Time of flight for the projectile: Since the projectile reaches highest
point after time t, the total time of flight will be double this time. Therefore,
the time of flight for the projectile is 2t= 2 ·1.02 = 2.04 s.
Step 4: Horizontal distance traveled by the projectile: The horizontal dis-
tance traveled by the projectile can be found using the horizontal compo-
nent of velocity and the time of flight. The horizontal distance is given by:
X=V0x·tSubstitute V0x= 10√3 m/s and t= 2.04 s to solve for X:
X= 10√3·2.04 = 20.4√3 meters
Step 5: Speed of the fragment moving along the original path after explosion:
Since the vertical component of velocity of the projectile immediately drops to
zero after explosion, the horizontal component will remain unchanged. Hence,
the speed of the fragment moving along the original path after the explosion is
equal to the horizontal component of velocity, which is 10√3 m/s.
Question 34
Question
A particle moves in the xy plane such that its position is given by r(t) =
(4t2+ 2t)i+ (3t−t3)j, where tis in seconds and the position components are
in meters. Find the magnitude of the velocity of the particle at t= 2 s.
Solution
Step 1: Find the velocity function v(t) by taking the derivative of the position
function r(t).
Step 1: v(t) = dr(t)
dt =d(4t2+ 2t)
dt i+d(3t−t3)
dt j
Step 2: Calculate the derivatives of the components of r(t).
Step 2: v(t) = (8t+ 2)i+ (3 −3t2)j
Step 3: Find the magnitude of the velocity at t= 2 s.
Step 3: |v(2)|=p(8(2) + 2)2+ (3 −3(2)2)2
Step 4: Simplify the expression and calculate the final answer.
Step 4: |v(2)|=p(16 + 2)2+ (3 −12)2=p182+ 92=√324 + 81 = √405 = 3√45 = 3√5 m/s
Therefore, the magnitude of the velocity of the particle at t= 2 s is 3√5
m/s.
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Question 35
Question
A projectile is launched with an initial speed of 50 m/s at an angle of 30◦above
the horizontal. Find the time it takes for the projectile to reach its maximum
height.
Solution
Step 1: Break the initial velocity into its horizontal and vertical components.
Let’s denote the initial velocity of 50 m/s as v0and the angle of 30◦as θ. The
horizontal component of velocity (v0x) is given by:
v0x=v0cos(θ)
Plugging in the values, we get:
v0x= 50 cos(30◦)
v0x= 50 ×√3
2
v0x= 25√3 m/s
The vertical component of velocity (v0y) is given by:
v0y=v0sin(θ)
Plugging in the values, we get:
v0y= 50 sin(30◦)
v0y= 50 ×1
2
v0y= 25 m/s
Step 2: Calculate the time it takes for the projectile to reach its maximum
height. The time taken to reach the maximum height can be calculated using
the vertical component of velocity:
vf=v0y−g·t
At the maximum height, the vertical velocity is 0 m/s (vf= 0). Therefore, we
can solve for the time t:
0 = 25 −9.8·t
9.8t= 25
t=25
9.8
t≈2.55 s
Therefore, the time it takes for the projectile to reach its maximum height
is approximately 2.55 seconds.
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