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 <u>vertical </u>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
