AP Physics
Chapter 4 Review
1. A particle starts from the origin at t = 0 with a
velocity of 6.0i secm and moves in the xy plane with a
constant acceleration of (-2.0i + 4.0j) secm .
At the
instant the particle achieves its maximum positive x
coordinate,
€ how far is it from the origin?
2
€
x = (6i) t + 12 (−2i )t 2
dx
v=
= 6 − 2t
dt
t = 3sec
x = (6i)t + 12 (−2i)t 2
x = (6
m
sec
x = 9iˆm
€
)3sec− 9m
y = 18 ˆjm
€
2
(a) t
2
1
y = 2 (4 )(3sec)
y=
1
2
1. A particle starts from the origin at t = 0 with a
velocity of 6.0i secm and moves in the xy plane with a
constant acceleration of (-2.0i + 4.0j) secm .
At the
instant the particle achieves its maximum positive x
coordinate,
€ how far is it from the origin?
2
€
r = 9iˆ + 18 ˆj m
(
2
2
ˆ
ˆ
r= i + j
r = 20.1m
€
)
2. A particle moves in the xy plane with a constant
acceleration given by a = −4.0 j secm2 . At t = 0 its
m
position and velocity are 10i meters and (−2i + 8 j) sec
,
respectively. What is the distance from the origin to
the particle at t = 2.0 s?
€
2
y = v iy t + 12 ay t €
y = (8
y = 8 ˆjm
x = xi + vxt
m
x =10m + (−2 sec
)(2sec)
x = 6iˆ m
m
sec
€
)(2sec) +
1
2
(−4 )(2sec)
m
sec2
2
2. A particle moves in the xy plane with a constant
acceleration given by a = −4.0 j secm2 . At t = 0 its
m
position and velocity are 10i meters and (−2i + 8 j) sec
,
respectively. What is the distance from the origin to
the particle at t = 2.0 s?
€
ˆ
ˆ€
(
)
r = 6i + 8 j m
r = iˆ 2 + ˆj 2
r = 10m
€
m
3. The initial speed of a cannon ball is 200 sec
. If the ball
is to strike a target that is at a horizontal distance of
3.0 Km from the cannon, what is the minimum time of
flight for the ball?
€
m
200 sec
3000m
€
€
v o2 sin2θ
R=
g


−1 Rg
2θ = sin  2 
 vo 
θ = 23.7°
m
3. The initial speed of a cannon ball is 200 sec
. If the ball
is to strike a target that is at a horizontal distance of
3.0 Km from the cannon, what is the minimum time of
flight for the ball?
€
v iy = v o sin θ
m
v iy = 80.4 sec
v fy = v iy + gt
v fy − v iy
m
−9.8 sec
2
€
=t
m
−80.4 sec
− 80.4
m
−9.8 sec
2
16.4 sec = t
m
sec
=t
4. A ball is thrown horizontally from the top of a building 100 m
high. The ball strikes the ground at a point 65 m horizontally
away from and below the point of release. What is the speed
of the ball just before it strikes the ground?
2y
t=
= 4.52sec
g
(
v = 14.4 iˆ − 44.3 ˆj
x
65m
m
vx = =
= 14.4 sec
t 4.52sec
v y = gt = −9.8 secm2 ( 4.52sec)
(
)
m
v y = −44.3 sec
€
)
v = iˆ 2 + ˆj 2
v = 14.4 2 + 44.32
m
v = 46.5 sec
m
sec
5. A baseball is hit at ground level. The ball is observed to
reach its maximum height above ground level 3.0 sec after
being hit. And 2.5 sec after reaching this maximum height,
the ball is observed to barely clear a fence that is 97.5 m
from where it was hit. How high is the fence?
Δy = v iy t + 12 gt 2
v fy = v iy + gt
0 = v iy + gt
v iy = gt = 9.8 secm 2 ( 3sec)
m
v iy = 29.4 sec
5. A baseball is hit at ground level. The ball is observed to
reach its maximum height above ground level 3.0 sec after
being hit. And 2.5 sec after reaching this maximum height,
the ball is observed to barely clear a fence that is 97.5 m
from where it was hit. How high is the fence?
Δy = v iy t + 12 gt 2
Δy = 29.4
m
sec
(5.5sec) − 4.9 (5.5sec)
Δy =161.7m −148.2m
Δy =13.4m
€
m
sec2
2
6. A rock is projected from the edge of the top of a building
m
with an initial velocity of 25 sec
at an angle of 53° above the
horizontal. The rock strikes the ground a horizontal distance
of 75 m from the base of the building. Assume that the
ground is level and that the side of the building is vertical.
€
How tall is the building?
m
v i = 25 sec
at 53°
m
25 sec
53°
€
€
€
v i = 15iˆ + 20 ˆj
(
)
m
sec
Δx 75m
t=
= m = 5 sec
v x 15 sec
6. A rock is projected from the edge of the top of a building
m
with an initial velocity of 25 sec
at an angle of 53° above the
horizontal. The rock strikes the ground a horizontal distance
of 75 m from the base of the building. Assume that the
ground is level and that the side of the building is vertical.
€
How tall is the building?
1
2
Δy = v iy t + gt
Δy = (20
m
sec
)(5sec) +
Δy = 22.5m
€
2
1
2
(−9.8 )(5sec)
m
sec 2
2
€
7.
A rock is projected from the edge of the top of a 100-ft tall
building at some unknown angle above the horizontal. The
rock strikes the ground a horizontal distance of 160 ft from
the base of the building 5.0 seconds after being projected.
Assume that the ground is level and that the side of the
building is vertical. Determine the speed with which the
rock was projected.
Δy = −100 ft
Δx =160 ft
160 ft
ft
vx =
= 32 sec
5sec
t = 5sec
1
2
Δy = v iy t + gt
€
Solve for
v iy .
2
1
2
Δy = v iy t + gt
2
(
ft
)
−100 ft + 32 sec2 (5sec)
Δy − 12 gt 2
= viy =
t
5sec
ft
v iy = 60 sec
1
2
€
v i = 32iˆ + 60 ˆj
(
)
ft
sec
= 68
ft
sec
2
8.
Wiley Coyote has missed the elusive roadrunner once
m
again. He leaves the edge of the cliff at 50 sec
horizontal
velocity. If the canyon is 250 m deep, how far from the
edge of the cliff does the coyote land?
€
2(250m)
2y
t=
=
= 7.14sec
m
g
9.8 sec2
m
Δx = v x t = (50 sec
)(7.14 sec)
€
€
Δx = 357m
€
9.
m
An artillery shell is fired with an initial velocity of 300 sec
at 53° above the horizontal. It explodes on a mountainside
42 seconds after firing. If x is horizontal and y vertical,
find the (x, y) coordinates where the shell explodes.
v i = 300
m
sec
Δx = v x t
at 53°
v i = 180iˆ + 240 ˆj
(
1
2
)
Δy = v iy t + gt
Δy = (240
€
m
sec
m
Δx = (180 sec
)(42sec)
Δx = 7560iˆ m
2
)( 42sec) + (−4.9
m€
sec
Δy = 1436.4 ˆj m
m
sec 2
)(42sec)
2
r = 7560iˆ + 1436.4 ˆj m
(
)
10.
€
In a CRT-TV set, an electron beam moves horizontally at
m
3.0 ×10 7 sec
across the cathode ray tube and strikes the
screen, 42 cm away. How far does the electron beam fall
while traversing this distance?
Δx 0.42m
−8
t= =
=1.4 ×10 sec
7 m
vx 3×10 sec
1
2
2
(
Δy = gt = −4.9
€
€
Δy = 9.6 ×10−16 m
m
sec 2
)(1.4 ×10
−8
sec)
2
11. A football is thrown upward at a 30° angle to the
horizontal. To throw a 40.0-meter pass, what must be
the initial speed of the ball?
2
i
v sin2θ
R=
g
(
m
40m
9.8
(
)
sec2
Rg
= vi =
sin2θ
0.866
m
v i = 21.28 sec
€
)
12.
A rifle is aimed horizontally toward the center of a
target 100 m away, but the bullet strikes 10 cm
below the center. Calculate the velocity of the bullet
just as it emerges from the rifle.
g = −9.8 secm 2
Δy = −0.1m Δx =100m v iy = 0
2y
t=
=
g
€
2(0.1m)
(
9.8 secm2
)
= 0.142sec
Δx
100m
m
vx =
=
= 704 sec
Δt 0.142sec
€
13. A student in the front of a school bus tosses a
ball to another student in the back of the bus
while the bus is moving forward at constant
velocity. The speed of the ball as seen by a
stationary observer in the street:
a. is less than that observed inside the bus.
b. is the same as that observed inside the bus.
c. is greater than that observed inside the bus.
d. may be either greater or smaller than that
observed inside the bus.
e. cannot be determined without knowing the
magnitudes of the velocities.
14. A projectile starts at the coordinate origin, where
the displacement vector also originates. The
initial velocity, vo, makes an angle θo with the
horizontal where 0 < θο < 90°. At the instant
when the projectile is at the highest point of its
trajectory, the displacement, velocity and
acceleration vectors are r, v and a. Which
statement is true?
a. r is parallel to v.
b. r is perpendicular to v.
c. v is parallel to a.
d. v is perpendicular to a.
e. r is perpendicular to a.