Page 109 - Calculus Demystified

CHAPTER 3 Applications of the Derivative
96
When the body hits the ground, its height is 0. Thus we know that
2
0 = h(T ) = h(25/8) =−16 · (25/8) + h 0 .
We may solve for h 0 to obtain
625
h 0 = .
4
Thus all the information about our falling body is given by
625
2
h(t) =−16t + .
4
At time t = 0 we have
625
h(0) = .
4
Thus the initial height of the falling body is 625/4ft = 156.25 ft.
Notice that, in the process of solving the last example, and in the discussion
preceding it, we learned that h 0 represents the initial height of the falling body and
v 0 represents the initial velocity of the falling body. This information will be useful
in the other examples that we examine.

EXAMPLE 3.15
A body is thrown straight down with an initial velocity of 10 feet per second.
It strikes the ground in 12 seconds. What was the initial height?
SOLUTION
We know that v 0 =−10 and that h(12) = 0. This is the information that we
must exploit in solving the problem. Now
2
h(t) =−16t − 10t + h 0 .
Thus
2
0 = h(12) =−16 · 12 − 10 · 12 + h 0 .
We may solve for h 0 to obtain

h 0 = 2424 ft.
The initial height is 2424 feet.

You Try It: A body is thrown straight up with initial velocity 5 feet per second
from a height of 40 feet. After how many seconds will it hit the ground? What will
be its maximum height?