A rock is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 64.7 m high. The rock hits the ground 18.0 m from the base of the cliff. How would this distance change if the rock was thrown at 10.0 m/s?

Refer to the diagram shown.

Assume g = 9.8 m/s² and ignore air resistance

When the rock is launched from a height of 64.7 m,

u = 5.0 m/s, the horizontal velocity

v = 0, the initial vertical velocity

If the rock hits the ground 18.0 m from the base of the cliff, then the time of flight is

t = (18.0 m) / (5.0 m/s) = 3.6 s

The vertical distance traveled is

s = (1/2) * (9.8 m/s²) * (3.6 s) ² = 63.504 m

Because this distance is less than 64.7 m, ground level is slightly higher away from the base of the cliff. It is higher by

64.7 - 63.504 = 1.196 m

If the rock is thrown at 10 m/s, the time of flight remains the same because acceleration due to gravity is the same.

Therefore the horizontal distance traveled is

(10.0 m/s) * (3.6 s) = 36.0 m

Answer: The distance will be 36.0 m