Answer
See below
Work Step by Step
Given: $v=80 mph\\
h(0)=2\\
g=32\frac{feet}{sec^2}$
and the distance of the wall is $30$ feet.
For the height of the ball:
$\frac{dh}{dt}=-gt\\
dh=-gt dt$
Integrate both sides:
$\int dh=\int -gt dt\\
h=-g.\int t dt$
The solution is $h(t)=-\frac{1}{2}gt^2+c$
Substitute $h(0)=2 \rightarrow h(0)=-\frac{1}{2}g(0)^2+c\\
\rightarrow c-0=2\\
\rightarrow c=2$
then $h(t)=2-\frac{1}{2}gt^2$
When the ball hits the ground, we have:
$h(t)=0\\
\rightarrow 2-\frac{1}{2}gt^2=0\\
\rightarrow \frac{1}{2}gt^2=2\\
\rightarrow gt^2=4\\
\rightarrow t^2=\frac{4}{g}\\
\rightarrow t=\sqrt \frac{4}{g}\\
\rightarrow t=\sqrt \frac{4}{32}=\sqrt \frac{1}{8}=\frac{1}{2\sqrt 2}$
The distance the ball travels before it hits the ground is:
$d=v.t\\
d=80.\frac{1}{2\sqrt 2}\\
d=80.\frac{5280}{3600}.\frac{1}{2\sqrt 2}=41.5$ feet
We have the length of the court is 40 feet, and since the ball travels 41.5 feet, it will hit the front wall before it falls to the ground.
