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?

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