Chapter 3.4, Problem 63E

To determine

To calculate: The time for which the opposing team has to return the ball before it bounces the court.

Answer to Problem 63E

The required time for which the opposing team has to return the ball before it bounces the court is 3.61 seconds.

Explanation of Solution

Given information:

The initial height of the ball = 10 feet

The initial vertical velocity = 55 feet per second

Formula used:

Quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Consider the required time be t , the initial height be $h_0$ and the initial vertical velocity be $v_0$ .

So, $h_0=10$ , $v_0=55$ and $h=0$

The model of the throwing of the ball can be written as:

  $h=-16t^2+v_{0}t+h_0$

Substitute 55 for $v_0$ , 10 for $h_0$ and 0 for h in the above equation and simplify for t .

  $\begin{array}{c}0=-16t^2+55t+10\\ 16t^2-55t-10=0\end{array}$

To solve the above equation for t , use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Substitute −55 for b , 16 for a and −10 for c in the above formula.

  $\begin{array}{c}t=\frac{-\left(-55\right)\pm \sqrt{{\left(-55\right)}^2-4\left(16\right)\left(-10\right)}}{2\left(16\right)}\\ =\frac{55\pm \sqrt{3025+640}}{32}\\ =\frac{55\pm \sqrt{3665}}{32}\\ =\frac{55\pm 60.54}{32}\end{array}$

Solve for t further.

  $\begin{array}{c}t=\frac{55+60.54}{32}\\ =\frac{115.54}{32}\\ =3.61\end{array}$

And, $\begin{array}{c}t=\frac{55-60.54}{32}\\ =-\frac{5.54}{32}\\ =-0.17\end{array}$

The value of t cannot be negative.

Hence, the required time for which the opposing team has to return the ball before it bounces the court is 3.61 seconds.