Chapter 4.5, Problem 59PPSE

Solution

a.

To Find: The height the ball goes in the air.

The maximum height the ball goes in the air is at $2.45\text{m}$ .

Given: $h\left(t\right)=-5t^2+7t$

Explanation:

Find the height the ball goes in the air:

The function $h\left(t\right)=-5t^2+7t$ is quadratic, so the maximum height occurs at the vertex:

The vertex is:

  $x=-\frac{b}{2a}$ for a quadratic equation $a{x}^2+bx+c=0$ .

Find the vertex of $-5t^2+7t=0$ :

  $\begin{array}{c}t=-\frac{7}{2\left(-5\right)}\\ =\frac{7}{10}\\ =0.7\end{array}$

Thus, the maximum height occurs at $t=0.7$

To find the height of the ball at $t=0.7$ , substitute $t$ value in $h\left(t\right)=-5t^2+7t$ :

  $\begin{array}{c}h\left(t\right)=-5{\left(0.7\right)}^2+7\left(0.7\right)\\ =-5\left(0.49\right)+4.9\\ =-2.45+4.9\\ =2.45\end{array}$

Thus, the maximum height the ball goes in the air is at $2.45\text{m}$ .

b.

To Find: The number of seconds the ball in the air between the toss and the serve when it $0.6\text{m}$ .

The ball is in air and hit of bat is $1.31\text{sec}$ .

Given: $h\left(t\right)=-5t^2+7t$

Explanation:

Plug $h=0.6$ solve for $t$ to find the number of seconds ball is in air and hit of ball :

Substitute $h=0.6$ in $h\left(t\right)=-5t^2+7t$ :

  $\begin{array}{c}0.6=-5t^2+7t\\ 5t^2-7t+0.6=0\end{array}$

To find the roots using the quadratic formula:

  $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=5,b=-7$ and $c=0.6$

  $\begin{array}{c}t=\frac{-\left(-7\right)\pm \sqrt{{\left(-7\right)}^2-4\left(5\right)\left(0.6\right)}}{2\left(5\right)}\\ =\frac{7\pm \sqrt{49-12}}{10}\\ =\frac{7\pm \sqrt{37}}{10}\\ =\frac{7\pm 6.08}{10}\\ =\frac{7-6.08}{10},\frac{7+6.08}{10}\\ =0.092,1.31\end{array}$

At this point, the ball is on its way up since $0.09$ seconds is very close to $0$ seconds. Meanwhile, the ball is falling at $1.31$ seconds. So, $1.31$ sec is more appropriate

Hence, the ball is in air and hit of bat is $1.31\text{sec}$ .