Chapter 1.9, Problem 7MRPS

Solution

a.

To find: The function with height $h$ (in feet).

  $h=-16t^2+50t+5$

Given information:

The ball through upward at height 5 feet with vertical velocity of 50 feet per second.

Calculations:

The path of ball is parabolic. Let the path equation be $h=-\frac{1}{2}g{t}^2+v_{0}t+h_0$

Initial height of the ball is, $h_o=5$

Initial velocity, $v_o=50$

Gravity, $g=32$

Substitute the value into the model,

  $\begin{array}{l}h=-\frac{1}{2}\left(32\right)t^2+50t+5\\ h=-16t^2+50t+5\end{array}$

Thus, the model the path is $h=-16t^2+50t+5$ .

b.

To find: The time of flight of ball.

  $3.222$ seconds.

Given information:

The ball through upward at height 5 feet with vertical velocity of 50 feet per second.

Calculations:

The model the path is $h=-16t^2+50t+5$ .

When ball hit the ground from the maximum height.

Substitute $h=0$ ,

  $\begin{array}{c}-16t^2+50t+5=0\\ t=-0.097,3.222\end{array}$

Time cannot be negative. So, the time of flight of ball is 3.222 seconds.

c.

To find: The maximum height of the ball.

  $44.063$ feet.

Given information:

The ball through upward at height 5 feet with vertical velocity of 50 feet per second.

Calculations:

The model the path is $h=-16t^2+50t+5$ .

Three methods to find the maximum height:

• Vertex form
• Energy conservation
• Derivative method

First, method to find the maximum height. As it is parabolic path and the maximum at vertex.

The vertex of the path is $\left(1.563,44.063\right)$ .

Thus, the maximum height is at 44.063 feet.

Energy conservation method: Total energy of the system is conserved.

  $\begin{array}{c}g{h}_0+\frac{1}{2}{v}_0^2=g{h}_{\max }\\ \left(32\right)\left(5\right)+\frac{1}{2}{\left(50\right)}^2=32h_{\max }\\ h_{\max }=\frac{1410}{32}\\ h_{\max }\approx 44.063\end{array}$

Thus, the maximum height is 44.063 feet.

Find the maximum height using derivative method:

As the derivative is zero at maximum height.

  $\begin{array}{c}h=-16t^2+50t+5\\ \frac{dh}{dt}=-32t+50\end{array}$

For maximum height, $\frac{dh}{dt}=0$

  $\begin{array}{c}-32t+50=0\\ t=1.5663\end{array}$

Substitute $t=1.5663$ and find out maximum height.

  $\begin{array}{c}h=-16{\left(1.5663\right)}^2+50\left(1.5663\right)+5\\ h_{\max }=44.063\end{array}$

Thus, the maximum height is 44.063.