Homework 4
Practice Problems
Section 2.7, 3.1 and 3.4
All of these problems contain multiple steps. Remember that the equations of motion applies to
two points of interest connect by a constant acceleration, and every two points that you chose can
produce two more equations of motion.
Question 1: 2 Points of Interest
A stone is dropped from the root of a building. A person in the building observes that the stone
crosses his window that is 2 meters tall in a time of 0.2 second. How far is the roof from the top of
the window?
Solution
There are three points here. I will put them all out here. The point on the roof is the origin.
roof to top
top to bottom
–9.8 m/s2
–9.8 m/s2
0s
ttop
ttop
ttop+0.2
0m
y
y
y–2 m
0 m/s
vtop
vtop
vbot
I will find the speed at the top of the window first through the window.
1
Δy = vi,y Δt − gΔt 2
2
⇒
− 2 = vi,y (0.2) − 4.9(0.2)2
⇒ vi,y = −9.02 m/s
Now comes the distance y.
Δ(vy2 ) = −2gΔy
⇒ (−9.02)2 = −2(9.8)y
⇒ y = −4.15 m
The distance from the roof to the top of the window is 4.15 m.
page 1
Question 2: Two Accelerations
A robot is dropped from a height of 50 meters. Some time after the drop, it opens a parachute that
drops its speed to a constant 0.5 m/s. How high should the parachute open if the robot is to land
20 s from the drop?
Solution
Here are the two parts of the robot’s trip. The origin is at the ground.
drop to parachute
parachute to ground
–9.8 m/s2
0 m/s2
0s
t
t
20 s
50 m
y
y
0m
0 m/s
v
–0.5 m/s
–0.5 m/s
In the first part of the trip,
1
Δy = vi,y Δt − gΔt 2
2
⇒ y − 50 = −4.9(t)2
In the second part of the trip,
1
Δy = vi,y Δt − gΔt 2
2
⇒
− y = −0.5(20 − t)
Together,
y = 50 − 4.9(t)2 = 0.5(20 − t) ⇒ 100 − 9.8t 2 = (20 − t)
9.8t 2 − t − 80 = 0
t = 2.9086 s
The height is
y = 0.5(20 − 2.9086) = 8.55 m
page 2
Question 3: Two Objects
Car A is 200 meters from the end of a straight track. Car B is 50 meters closer to the end of the
track. Both cars start to accelerate at the same time. Car A accelerates at a rate of 2.5 m/s2.
What acceleration must car B have in order to arrive at the end of the track at the same time as
car A?
Solution
I will place the origin at the initial position of car A. Car B will be at 50 meters. Here are the
information for the motion of the cars.
Car A
Car B
2.5 m/s2
a
0s
t
0s
t
0
200 m
50 m
200 m
0 m/s
vA
0 m/s
vB
Car A does this.
1
Δy = vi,y Δt + aΔt 2
2
⇒
1
200 = (2.5)(t)2
2
Car B does this.
1
Δy = vi,y Δt aΔt 2
2
1
⇒ 150 = a(t)2
2
Car A says that the total elapsed time is
t 2 = 160
Car B says the acceleration is
1
300
150 = a(160) ⇒ a =
= 1.875 m/s 2
2
160
page 3
Question 4: Projectile Motion
A ball is kicked from the edge of a 25 meter cliff. The speed of the ball is 10 m/s and the angle is
30° above the horizontal direction. How much time does it take the ball to reach to the height of
10 meters above the ground level below?
10 m/s
30°
25 m
10 m
Solution
This is a projectile motion problem. I will use the base of the cliff as the origin.
x direction
y direction
0 m/s2
–g m/s
0s
t
0s
t
0m
x
25 m
10 m
10cos30°
vA
10sin30°
vB
Using the y direction,
1
Δy = vi,y Δt aΔt 2
2
1
⇒ 10 − 25 = 10 sin 30°t − (9.8)t 2
2
4.9t 2 − 5t − 15 = 0
⇒ t = 2.3327 and − 1.3123 s
⇒
− 15 = 5t − 4.9t 2
The correct answer is positive so it is 2.33 s.
page 4
Question 5: Two Dimensional Motion
A rectangular table has a width of 1 meter and a length of 2 meters. The table is not level. When
a ball is released at the corner of the table, the ball rolls off the table in the direction shown over a
time of 3 seconds. The acceleration of the ball is constant.
Here is the view from above the table.
2m
60°
1m
3s
Let the length of the table be the x direction and right be positive and width be the y direction and
down be positive.
a. What is the acceleration vector of the ball?
b. What is the speed of the ball when it rolls off of the table?
Now, you roll the ball in the x direction instead of just letting it go. The ball rolls off the table at the
opposite corner of the table. Here is what the trajectory looks like.
vi
2m
1m
c. What is the acceleration vector of the ball?
d. How much time does the ball spend on the table in this case?
e. What is the velocity of the ball vi after you released it?
page 5
Solution
a. The ball rolls spontaneously in the y direction so there is an acceleration in the y direction. It
travels a distance of 1 meter in 3 seconds under the constant acceleration so
1
Δy = ay Δt 2
2
1
⇒ 1 = ay (3)2
2
⇒ ay =
2
m/s 2 = 0.222 m/s 2
9
The ball also rolls spontaneously in the x direction so there is an acceleration in the x direction
too.
1
Δx = ax Δt 2
2
⇒
1
1
= ax (3)2
tan 60° 2
⇒ ax = 0.1283 m/s 2
The acceleration vector is
!
a = (0.1283 m/s 2 )(x̂) + (0.222 m/s 2 )(ŷ)
b. The y direction final speed is
!
!
vy = ay Δt ⇒ vy = (0.2222)(3) = (0.6667 m/s)(x̂)
The x direction final speed is
!
!
vx = ax Δt ⇒ vx = (0.1283)(3) = (0.3849 m/s)(x̂)
The speed is
!
v =
!
!
vx2 + vy2 = (0.3849 m/s)2 + (0.6667 m/s)2 = 0.770 m/s
c. Since there is no change to the table, the acceleration is the same as before.
d. The motion in the y direction is the same as before so it still spends 3 seconds on the table.
e. In the x direction, the motion is described by this.
1
Δx = vi Δt + aΔt 2
2
⇒
1
2 = vi (3) + (0.1283)(3)2
2
⇒ vi = 0.47422 m/s
page 6