Question

A diver running 1.6 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.0 s later. How high was the cliff

Answer 1

Step-by-step explanation:

Using the equations of motion;

S = ut+1/2gt²

u is the velocity

g is the acceleration due to gravity

t is the time taken

Given

u = 1.6m/s

g = 9.81m/s²

t = 3.0secs

S is the height of the cliff

Substitute

S = 1.6(3)+1/2(9.81)(3)²

S = 4.8+44.145

S = 48.945m

Hence the cliff is 48.945m high