Chapter 8.4, Problem 29PPS

(a)

To determine

The equation that models Jerald’s free throw.

Answer to Problem 29PPS

  $10=-16t^2+20t+6$

Explanation of Solution

Given:

The height of the ball is 6 feet.

The initial velocity is 20feet per second.

Calculation:

The solution for the given statement can be obtained as:

  $h=-16t^2+vt+h0$

Therefore,

  $\begin{array}{l}h=-16t^2+vt+h0\\ 10=-16t^2+20t+6\end{array}$

(b)

To determine

The time taken by the ball to reach the hoop.

Answer to Problem 29PPS

  $t=1$

Explanation of Solution

Given:

The height of the ball is 6 feet.

The initial velocity is 20feet per second.

Calculation:

The solution for the given statement can be obtained as:

  $\begin{array}{l}10=-16t^2+20t+6\\ 0=-16t^2+20t-4\\ 0=-4(4t^2-5t+1)\\ 0=4\left[\left(4t^2-4t\right)-(t-1)\right]\\ 0=4\left[4t\left(t-1\right)-\left(t-1\right)\right]\\ 0=4(t-1)(t-1)\\ 4t-1=0,t=\frac{1}{4}\\ or,\\ t-1=0,t=1\end{array}$

(c)

To determine

Whether the ball is in the air for more or less time.

Answer to Problem 29PPS

  $\text{Less time}\text{.}$

Explanation of Solution

Given:

The height of the ball is 6 feet.

The initial velocity is 20feet per second.

Calculation:

The ball will be in the air for less time.

  $\therefore \text{ The ball is 6 seconds in the air}\text{.}$