Answer
See below
Work Step by Step
From part a, we have: $h=5+50t-16t^2$
The ball reaches the ground when $h=0$, so:
$-16t^2+50t+5=0$
Given: $a=-16\\b=50\\c=2$
The solution of the equation is: $t=\frac{-50\pm \sqrt 50^2-4(-16)(2)}{2(-16)}=\frac{-50\pm \sqrt 2628}{-32}=\frac{50\pm \sqrt 2628}{32}\\
\rightarrow t\approx3.165 \lor t\approx-0.0395$
Time cannot be negative, so the solution to the problem is $t\approx3.165$
Hence, the ball reaches the ground after 3.165 seconds.
