Written Homework 7 Solutions

(a) Find a formula for the velocity v = f ′ (t) of the ball after t seconds. Check that your
formula agrees with the given information that the initial velocity of the ball is 0
feet/second.
Solution: Since we’re given that v = f ′ (t) and f(t) = −16t 2 + 200, it’s easy to see
that v(t) = −32t. Now we can check that v0 = 0: v(0) = −32 · 0 = 0.
(b) Draw graphs of both the velocity and the distance as functions of time. What time
interval makes physical sense in this situation? (For example, does t < 0 make sense?
Does the distance formula make sense after the ball hits the ground?)
Solution: The time interval that makes the most sense is 0 ≤ t ≤ 4 since we think
of time starting at zero and the ball hits the ground shortly before t = 4.
160
80
0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4
-80
-160
y = -16x² + 200
y = -32x
(c) At what time does the ball hit the ground? What is its velocity then?
Solution: The ball hits the ground when d = f(t) = −16t 2 +200 = 0 which happens
when t ≈ 3.54. When t = 3.54, the ball’s velocity is v(3.54) = −32 · 3.54 = −113.14.
18. A second ball is tossed straight up from the top of the same building with a velocity of 10
feet per second. After t seconds have passed, the distance from the ground to the ball is
d = f(t) = −16t 2 + 10t + 200 feet.
(a) Find a formula for the velocity of the second ball. Does the formula agree with given
information that the initial velocity is +10 feet per second? Compare the velocity
formulas for the two balls; how are they similar, and how are they different?
Solution: Again, we know f(t) and that v = f ′ (t), so we get that v = −32t + 10.
(b) Draw graphs of both the velocity and the distance as functions of time. What time
interval makes physical sense in this situation?
Solution: Again, the time interval that makes the most sense is 0 ≤ t ≤ 4.
4