Chapter 11.2, Problem 73STP

To determine

Find the time the soccer ball will be in air.

Answer to Problem 73STP

The soccer ball will be in the air for 2.26 sec.

Explanation of Solution

Given:

A soccer ball is kicked with an initial upward velocity of 35 ft/ sec from a starting height of 2.5 ft. If no one touches the ball, how long will it be in the air? Use the formula $h=-16t^2+vt+c$ where h is the ball`s height at a time t, v is the initial upward velocity, and c is the starting height. Show your work.

Concept Used:

Use the vertical motion formula $h=-16t^2+vt+c$ . A soccer ball is kicked with a starting upward velocity of 35 ft/ sec from a starting height of 2.5 ft. Substitute the values into the vertical motion formula.

To find the time the ball will be in the air , substitute h = 0 and solve for the time from the vertical motion formula.

Calculation:

The starting velocity is 35 ft/sec, when the ball is 2.5 ft high.

So, $v\text{ = }35,h=2.5\text{ and }t=0$

  $\begin{array}{l}h=-16t^2+vt+c\\ $ 2.5=-16{(0)}^2+35(0)+c$c=2.5\end{array}$

So, the vertical motion formula will be: $h=-16t^2+35t+2.5$

To find the soccer ball will be in the air , substitute h = 0

Let h = 0 $\Rightarrow -16t^2+35t+2.5=0$

Now solve for the time t

  $\begin{array}{l}\text{To}\text{slove}\text{for}t,\text{use}\text{quadratic}\text{formula}\\ a{t}^2+bt+c=0,\text{then}t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ -16t^2+35t+2.5=0\\ a=-16;b=35;c=2.5\\ t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ t=\frac{-35\pm \sqrt{{35}^2-4·-16·2.5}}{2·-16}=\frac{-35\pm \sqrt{1225+160}}{-32}=\frac{-35\pm \sqrt{1385}}{-32}\\ t=\frac{-35+\sqrt{1385}}{-32}\text{and}\frac{-35-\sqrt{1385}}{-32}\\ t=-0.06923\text{and}2.25673\\ t=2.26\sec \end{array}$

Time cannot be negative, so the soccer ball will be in the air for 2.26 sec.

Thus, the soccer ball will be in the air for 2.26 sec.