PHY 110
Fall 2016
Quiz 2
Solution
1. A rock is thrown downward from the top of a building with a speed of 50 ft/s. If it strikes the
ground 3 s later, how tall is the building?
a) 180 ft
b) 230 ft
c) 265 ft
d) 295 ft
Set x = 0 at the top of the building, with positive downward. Then both v0 and a are positive
1
numbers and the distance equation becomes x = 0 + 50(3) + ( 32.2 ) (3)2 = 294.9 ft
2
2. A plane flying horizontally at a constant speed of 300 ft/s crosses the starting line of a
racetrack just as a motorcycle starts the race from rest. If the racetrack is 2500 ft long, what
constant acceleration must the motorcycle achieve to reach the end of the track at the same time
the plane gets there?
a) 12 ft/s2
b) 35 ft/s2
c) 54 ft/s2
d) 72 ft/s2
2500
= 8.3 s . Then the distance equation
300
2
1
ft
for the motorcycle is 2500 = 0 + 0 + a ( 8.3 ) , which gives a = 72 2
2
s
Use the plane to find the total time for the race: ttotal =
3. A ball is dropped from the top of a 200 ft tower at the same time another ball is launched
upward from the ground. If the two balls meet exactly half way up the building, what was the
initial velocity of the ball launched from the ground?
a) 63.9 ft/s
b) 80.2 ft/s
c) 107 ft/s
d) 125 ft/s
Use the dropped ball to set the time. Set x = 0 at the top, positive downward for this ball. The
1
distance equation becomes 100 = 0 + 0 + ( 32.2 ) t 2 , which gives t = 2.49222 s. Now set x = 0 at
2
the bottom, positive upward for the ball thrown upward from the ground. The distance equation
1
ft
2
for this ball becomes 100 = 0 + v0 ( 2.49222 ) + ( −32.2 ) ( 2.49222 ) , which gives v0 = 80.2
2
s
4. A sports car and a large truck are traveling side by side at the same speed and both apply the
brakes at the same instant. The braking deceleration of the sports car is twice the braking
deceleration of the truck. If the car travels a distance dcar before stopping and the truck travels a
d
distance dtruck before stopping, what is the ratio truck ?
dcar
a)
2
b) 2
c) 4
d) 16
Convince yourself that the stopping equation for each vehicle becomes v02 = 2ad , where a and d
refer to the same vehicle, either the car or the truck. Since v0 is the same for both vehicles, this
means that acar dcar = atruck dtruck . So
dtruck acar
=
=2.
dcar atruck
5. A ball is thrown straight down from the top of a building that is 40 m tall. How fast must it
be thrown so that its velocity at the ground is 50 m/s?
a) 41.4 m/s
b) 57.9 m/s
c) 63.2 m/s
d) 75.5 m/s
Since info about positions and velocities is given or required, use the stopping equation. Set x =
0 at the top, with positive downward.
50 2 − v02 = 2 ( 9.81) ( 40 − 0 ) , which gives v0 = 41.4
m
.
s
The acceleration due to gravity is 9.81 m/s2 or 32.2 ft/s2. Some possibly useful equations for 1dimensional motion:
aavg =
Δv
Δt
v = v0 + at
1
x = x0 + v0t + at 2
2
To solve Ax 2 + Bx + C = 0, x =
v 2 − v02 = 2a(x − x0 )
−B ± B 2 − 4AC
2A