On-line Math 21

2.4  Applications of the derivative.

Example 1 A ball is thrown vertically upward with a velocity of 80 ft/sec , and its position t seconds later is s(t) : = 80t-16t² feet up. How high does it get, and how fast is it moving when it hits the ground again?

Solution:

The key thing for the first question is to note that the ball reaches velocity 0 when it is at its highest point. Then, solve

0 = v(t) = 80-32t,

which implies that t = 80/32 = 5/2 . The maximum height is then

s(5/2) = 80(5/2)-16(5/2)2 = 200-100 = 100.

For the second question, solve for s = 0 :

0 = s(t) = 80t-16t2 = 16t(5-t),

so t = 0 or t = 5 . Since the motion started when t = 0 , the ball had to hit the ground when t = 5 . To then answer the question, the velocity is v(5) = 80-32·5 = -80. This makes sense, since it really means 80 ft/sec downward.

Copyright (c) 2000 by David L. Johnson.