Chapter 11.2, Problem 36E

To determine

Tofind:the maximum height of the ball.

Answer to Problem 36E

  $85\text{feet}$ .

Explanation of Solution

Given:

A basket-ball player popped a baseball straight up at an initial velocity of $72\text{feet/sec}$ .

The distance s above the ground after t seconds is described by $s=v_{o}t-16t^2+4$

Calculation:

According to the question,

The distance s above the ground after t seconds is described by

  $\begin{array}{l}s=v_{o}t-16t^2+4\\ s=72t-16t^2+4\end{array}$

Now, differentiate the above equation with respect to t. Then,

  $v=72-32t$

When the object reached out, the velocity at that time is 0. Thus,

  $\begin{array}{l}0=72-32t\{t=0\}\\ \Rightarrow 32t=72\\ \Rightarrow t=\frac{72}{32}=2.25\sec \end{array}$

Therefore,

  $\begin{array}{l}s=72t-16t^2+4\\ s=72\left(2.25\right)-16{\left(2.25\right)}^2+4\left\{\text{plug the value}\right\}\\ s=162-81+4=85\text{feet}\end{array}$

Hence, the maximum height of the ball is $85\text{feet}$ .