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17 October, 19:39

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?

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  1. 17 October, 22:04
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    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
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