Answer
The answer is below.
Work Step by Step
We know the following equation:
$v_{0y}t+\frac{1}{2}gt^2$
Simplifying this equation and knowing that the initial x-velocity of the second ball is 0, we find:
$y=h-\frac{g}{2v_0^2cos^2\theta}x^2$
We know that the angle the first ball is thrown at is 45 degrees, and we know that $x=\frac{h}{tan45}$, so it follows:
$y = h - \frac{g}{v_0^2}\times \frac{h^2}{1}$
$y = \frac{hv_0^2 - gh^2}{v_0^2}$
Simplifying, we find that $v_0$ must be at least:
$v_0=\sqrt{\frac{hv_0^2 - gh^2}{y}}$
b) Using parabolic motion, we find that the minimum speed is: $v_0=\sqrt{gh}$.
