Chapter 4.3, Problem 70PPS

(a)

To determine

To Find: The time for which the ball is in the air.

Answer to Problem 70PPS

The time for which the ball is in the air is $3\sec$ .

Explanation of Solution

Given:

The equation to model the height that the ball attains is $h\left(t\right)=-4.9t^2+14.7t$ and the model to determine the distance travelled by the ball is $d\left(t\right)=16t$ .

Calculation:

The ball is in the air when the time is more than 0 then the time for which the ball is in the air is,

$\begin{array}{l}-4.9t^2+14.7t=0\\ -t\left(4.9t-14.7\right)=0\\ t=0,3\end{array}$

Thus, the time for which the ball is in the air is $3\sec$ .

(b)

To determine

To Find: The distance which the ball travelled before it hit the ground.

Answer to Problem 70PPS

Thus, the distance travelled by the ball is $48$ units.

Explanation of Solution

Consider the model for the distance is,

$d\left(t\right)=16t$

Then, the distance travelled by the ball is,

$\begin{array}{c}16t=16.3\\ =48\end{array}$

(c)

To determine

To Find: The maximum height that the ball attains.

Answer to Problem 70PPS

The maximum height is $11.025$ .

Explanation of Solution

Consider the maximum height that the ball attains is the x coordinate of the vertex of the height function by the use of the axis of symmetry equation as,

$\begin{array}{c}x=\frac{-b}{2a}\\ =-\frac{14.7}{-9.8}\\ =\frac{3}{2}\end{array}$

Thus, the maximum height that the ball attains is ,

$\begin{array}{c}h\left(\frac{3}{2}\right)=-4.9\left(\frac{3}{2}\right)+14.7\left(\frac{3}{2}\right)\\ =11.025\end{array}$