An object is projected horizontally at 8.0 m/s from the top of a 122.5 m cliff. how far from the base of the cliff will the object strike the ground?

Refer to the diagram shown below.

Assume g = 9.8 m/s² and neglect air resistance.

All quantities are measured as positive upward.

The vertical launch velocity is u = 0.

Let t = the time of flight.

Use the formula s = ut + (1/2) at²

Then

(-122.5 m) = 0 + (1/2) * (-9.8 m/s²) * (t s) ²

t² = 122.5/4.9 = 25

t = 5 s

The horizontal distance traveled is

d = (8.0 m/s) * (5 s) = 40 m

Answer: 40 m