PHY 108
Quiz 2
Solution
The correct answer is shown in bold font.
1. You throw a ball vertically downward with an initial speed of 20 ft/s from the top of Watterson Towers (height = 200
ft). How fast is the ball moving when it hits the ground?
a) 79.2 ft/s
b) 115 ft/s
c) 238 ft/s
d) 467 ft/s
Choose x = 0 at the top of the tower with positive downward. Use the stopping/launching equation v 2 " v 02 = 2a( x " x 0 )
with v0 = 20, a = 32.2, x = 200, and x0 = 0.
2. Ignoring the effects of air resistance, the motion of a golf ball in flight can be described as
a)
b)
c)
d)
vertical motion with constant acceleration and horizontal motion with constant
velocity.
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vertical motion with constant velocity and horizontal motion with constant acceleration.
constant (non-zero) acceleration motion in both horizontal and vertical directions.
horizontal motion with positive acceleration and vertical motion with negative acceleration.
3. A satellite with an initial velocity of 2500 m/s enters a “forbidden region” in which aliens have imposed a force field
that repels the satellite directly along its line of flight with an acceleration of 400 m/s2. How long will the satellite remain
in the forbidden region? Careful about signs, and notice that the total time in the forbidden region involves time going in
and an equal time coming out.
a) 12.5 s
b) 34.9 s
c) 57.2 s
2500 m/s
d) 81.6 s
400 m/s2
Let x = 0 be at the beginning of the forbidden region, with positive as the initial direction of the satellite (to the right).
How much time will elapse until the satellite stops? Use v = v0 + at with v = 0, v0 = 2500 and a = –400 to get t = 6.25 s.
Double that to get the entire time spent in the forbidden region.
4. The roadway of the Golden Gate Bridge is about 240 ft above the water beneath it. Ignoring air resistance, how long
would it take a rock dropped from the roadway to reach the water?
a) 3.86 s
b) 4.13 s
c) 6.25 s
d) 9.72 s
Let x = 0 be at the water’s surface with positive upward. Then use x = x 0 + v 0 t +
a = –32.2 to find t = 3.86 s. You don’t need the quadratic formula, since v0 = 0.
1 2
at with x = 0, x0 = 240, and
2
5. A racecar goes from rest to a speed of 133 ft/s with constant acceleration. What is the car’s average speed during this
motion?
a) 41.7 ft/s
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b) 54.8 ft/s
c) 66.5 ft/s
d) 73.9 ft/s
Since the acceleration is constant (and a graph of v vs. t is a straight line), the average velocity is just the average value of
the height of a line, which in this case is just (vinitial + vfinal)/2 or (0 + 133)/2 = 66.5 ft/s.
For motion with constant acceleration, x = x 0 + v 0 t +
Acceleration due to gravity is 9.80 m/s2 or 32.2 ft/s2
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1 2
at
2
v = v 0 + at v 2 " v 02 = 2a( x " x 0 )
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