Question

A stone was dropped off a cliff and hit the ground with a speed of 104 ft/s. what is the height of the cliff? (use 32 ft/s2 for the acceleration due to gravity.)

Answer 1

Height of cliff= 169 ft

Explanation:

using kinematic equation:

Vf²= Vi²+ 2 g h

Vf= Final velocity= 104 ft/s

Vi= initial velocity=0

g= acceleration due to gravity=32 ft/s²

h= height of cliff

(104)²= 0 + 2 (32) h

64 h=10816

h=169 ft

Answer 2

Let the stone is dropped off a cliff of height S m ft.

Given, initial velocity of the stone, u= 0 ft/s

Final velocity of the stone, v= 104 ft/s

Acceleration of the stone= Acceleration due to gravity = 32 ft/s²

Using the third equation of motion,

v² - u² = 2aS

Substituting the values we get,

$(104)^{2}-(0)^{2}=2*32*S$

S= 169 ft

Height of the cliff is 169 ft.