Question

cliff divers at acapulco jump into the sea from a cliff 35 m high. At the level of the sea, a rock sticks out a horizontal distance of 10.2 m. The acceleration of gravity is 9.8m/s2. With what minumum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock?

Answer 1

$$x=vt\\h=0.5gt^2\\x=v\sqrt{\frac{2h}{g}}\\10.2=v\sqrt{\frac{2(35)}{9.8}}\\v=3.82\frac{m}{s}$$