Chapter 3.5, Problem 34AYU

(a)

To determine

To Calculate: The time $t$ where the ball strikes the ground.

Answer to Problem 34AYU

The time at which the throwing ball strikes the ground is $\text{}5\sec$ .

Explanation of Solution

Given Information:

  $\begin{array}{l}S=96t-16t^2\\ V=80ft/s\end{array}$

Calculation:

The ball strikes the ground that means the distance is zero. So, the time to touch the ground is calculated as:

  $\begin{array}{l}S=96t-16t^2\\ 0=96t-16t^2\\ 16\left(-t^2+6t\right)=0\\ t\left(-t+6\right)=0\\ t=6\sec \end{array}$

(b)

To determine

To Calculate:The time $t$ at which the ball more than $128$ feet above the ground.

Answer to Problem 34AYU

The time when the ball is more than $128$ feet above the ground is $2\sec$ or $4\sec$ .

Explanation of Solution

Given Information:

  $S=96t-16t^2$

Calculation:

The time is estimated when the ball s more than the height of the building from the ground as:

  $\begin{array}{l}S=96t-16t^2\\ 128<96t-16t^2\\ 96t-16t^2-128<0\\ 16\left(-t^2+6t-8\right)<0\\ -t^2+6t-8<0\\ t^2-6t+8<0\\ \left(t-2\right)\left(t-4\right)<0\\ t=2\sec OR\text{}t=4\sec \end{array}$