h = Vo*t + 0.5g*t^2 = 37.1 m.
0 + 4.9t^2 = 37.1
t^2 = 7.57
t = 2.75 s.
d = V*t = 13.9 m.
2.75V = 13.9
V = 5.05 m/s.
Cliff divers at Acapulco jump into the sea from a cliff 37.1 m high. At the level of the sea, a rock sticks out a horizontal distance of13.39 m. The acceleration of gravity is 9.8 m/s^2. With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock?
1 answer
1 answer