Question

A stone is thrown horizontally with an initial speed of 9 m/s from the edge of a cliff. A stop watch measures the stone's trajectory time from the top of the cliff to the bottom to be 4.7 s. What is the height of the cliff? Round to the nearest tenth of a meter

Answer 1

Answer:

the height of the cliff is equal to 108.241 m

Explanation:

given,

initial speed of the stone horizontally  = 9 m/s

vertical speed of the stone = 0 m/s

time taken of the trajectory = 4.7 s

using equation of motion

$s = ut + \dfrac{1}{2}at^2$

$s =0 \times 4.7 + \dfrac{1}{2} \times 9.8 \times 4.7^2$

$s = 0 + 4.9 \times 4.7^2$

$s = 4.9 \times 22.09$

s = 108.241 m

hence, the height of the cliff is equal to 108.241 m