Answer
a) $v_0>\sqrt{.5hg}$
b) $y = h -\frac{gh^2}{2v_0^2}$
Work Step by Step
a) We know the following equation for the value of y, recalling that $a=g$ in this case since the only acceleration is due to gravity:
$y = v_0t+\frac{1}{2}gt^2$
We also know:
$y=h-\frac{1}{2}gt^2$
Setting these equal gives:
$ v_0t+\frac{1}{2}gt^2=h-\frac{1}{2}gt^2$
$v_0t=h \\ v_0=\frac{h}{t}$
We know that the value of t is $t = \sqrt{\frac{2h}{g}}$, so we obtain:
$v_0>\frac{h}{\sqrt{\frac{2h}{g}}}$
$v_0>\sqrt{.5hg}$
b) We use the second equation, which is $y=h_0-\frac{1}{2}gt^2$. Using substitution, we find:
$y = h_0 -\frac{gh_0^2}{2v_0^2}$
