Chapter 9, Problem 13MCQ

To determine

Find the time need the ball in the air.

Answer to Problem 13MCQ

Explanation of Solution

Given:

Juan hits a base ball. The equation $h=-16t^2+120t$ models the height, in feer of the ball after t seconds. How long is the ball in the air?

.

Concept Used:

The ball is no longer in the air when it lands on the ground which occurs when h = 0, so substitute h = 0 into the given equation.

  $\begin{array}{l}h=-16t^2+120t+[\text{Original}\text{Equation}]\\ 0=-16t^2+120t\end{array}$

Calculation:

Solve for t by using quadratic formula.

  $\begin{array}{l}a{x}^2+bx+c=0,\text{then}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ h=-16t^2+120t[\text{Original}\text{Equation}]\\ 0=-16t^2+120t\\ a=-16;b=120;c=0\\ t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ t=\frac{-120\pm \sqrt{{(120)}^2-4·-16·0}}{2·-16}=\frac{-120\pm \sqrt{14400}}{-32}=\frac{-120\pm 120}{-32}\\ t=\frac{-120+120}{-32}\text{and}\frac{-120-120}{-32}\\ t=0\text{and}t=7.5\end{array}$

Time $t=0$ second corresponds to when the ball is hit, so $t=7.5$ seconds corresponds to when it lands on the ground.

The ball is then in the air for about 7.5 seconds.

Thus, the ball is then in the air for about 7.5 seconds.