Answer
When the ball's height is 5 feet, the horizontal distance traveled by the ball is more than 100 feet. Therefore the ball would clear the fence.
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 $y= 5$:
$y = 55.4~t-16~t^2+3$
$-16~t^2+55.4~t-2 = 0$
We can use the quadratic formula to find $t$:
$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-55.4 \pm \sqrt{(55.4)^2-(4)(-16)(-2)}}{(2)(-16)}$
$t = \frac{-55.4 \pm \sqrt{2941.16}}{-32}$
$t = 3.426~s, 0.036~s$
We can see that $t = 3.246~s$ is the time when the baseball is at a height of 5 feet and on its way down.
We can find the horizontal distance at this time:
$x = 32~t$
$x = (32~ft/s)(3.246~s)$
$x = 103.9~ft$
When the ball's height is 5 feet, the horizontal distance traveled by the ball is more than 100 feet. Therefore the ball would clear the fence.
