## Essential University Physics: Volume 1 (3rd Edition)

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: $v_0=\sqrt{gh}$.