Chapter 3.5, Problem 33AYU

(a)

To determine

To Calculate: Time $\text{}t$ will the ball strike the ground.

Answer to Problem 33AYU

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

Explanation of Solution

Given Information:

  

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

Calculation:

When a ball is thrown in upward direction at the top of the building of height $\left(h=96ft\right)$ with initial velocity of $\left(80ft/s\right)$ , the distance of the ball from the ground is $S=80t-16t^2$

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

  $\begin{array}{l}S=80t-16t^2\\ 0=80t-16t^2\\ 16\left(-t^2+5t\right)=0\\ t\left(-t+5\right)=0\\ t=5\sec \end{array}$

(b)

To determine

To Calculate: The time where $\text{}t$ is the ball more than $96$ feet above the ground.

Answer to Problem 33AYU

The time when the ball is more than given height of building $\left(96ft\right)$ above the ground is $\text{}6\sec$ .

Explanation of Solution

Given Information:

  

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

Calculation:

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

  $\begin{array}{l}S=80t-16t^2\\ 96<80t-16t^2\\ 80t-16t^2-96<0\\ 16\left(-t^2+5t+6\right)<0\\ -t^2+5t+6<0\\ t^2-5t-6<0\\ \left(t-6\right)\left(t+1\right)<0\\ t=-1\sec OR\text{}t=6\sec \end{array}$