Answer to Question #124625 in Physics for jn

Question

In a game war, one team sets base on a cliff 15m high and 60m away from the opponent’s base. At what velocity must the attack be launched so that the lower base will be hit? The initial launch is at 20 degrees below the horizontal?

Answer 1

$$0=h-v\sin{20}t-0.5gt^2\\0=15-v\sin{20}t-0.5(9.8)t^2$$

$$x=v\cos{20}t\to t=\frac{60}{v\cos{20}}$$

$$0=15-\frac{60}{\cos{20}}\sin{20}-0.5(9.8)\frac{60^2}{v^2\cos{20}^2}$$

$$v=54\frac{m}{s}$$