Answer
The horizontal distance traveled by the ball in the air is 112.5 feet
Work Step by Step
We can find an equation for the horizontal position $x$:
$x = (v~cos~\theta)~t$
$x = (64~cos~60^{\circ})~t$
$x = 32~t$
We can find an equation for the vertical position $y$:
$y = (v~sin~\theta)~t-16~t^2+h$
$y = (64~sin~60^{\circ})~t-16~t^2+3$
$y = 55.4~t-16~t^2+3$
We can find the time $t$ when $y= 0$:
$y = 55.4~t-16~t^2+3$
$-16~t^2+55.4~t+3 = 0$
We can use the quadratic formula to find $t$:
$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-55.4 \pm \sqrt{(55.4)^2-(4)(-16)(3)}}{(2)(-16)}$
$t = \frac{-55.4 \pm \sqrt{3261.16}}{-32}$
$t = 3.516~s, -0.053~s$
Since $t$ must be positive, the solution is $t = 3.516~s$
We can find the horizontal distance:
$x = 32~t$
$x = (32~ft/s)(3.516~s)$
$x = 112.5~ft$
The horizontal distance traveled by the ball in the air is 112.5 feet
