Question

A person standing at the edge of a seaside cliff kicks a stone horizontally over the edge with a speed of 18 m/s. The cliff is 52 meters above the water's surface. How long does it take for the stone to fall to the water?

Answer 1

Answer:

it would take 3.26 seg for the stone to fall to the water

Explanation:

If we ignore air friction then:

h=h₀ + v₀*t -1/2*g*t²

where

h= coordinates of the stone in the y axis ( height of the stone relative to the surface of the water )

h₀ = initial coordinates of the stone ( height of the cliff relative to the surface of the water = 52 m )

v₀ = initial vertical velocity = 0 ( since the ball is kicked horizontally , has only initial horizontal velocity , and has 0 vertical velocity )

t = time to reach a height h

g = gravity = 9.8 m/s²

since v₀ =0

h= h₀ - 1/2*g*t²

h₀ - h =  1/2*g*t²

t= √[2(h₀ - h)/g]

when the stone hits the ground h=0 ( height=0) , then replacing values

t=√[2(h₀ - h)/g]=√[2(52 m- 0 m )/(9.8m/s²)] = 3.26 seg

t= 3.26 seg

it would take 3.26 seg for the stone to fall to the water