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1 July, 08:35

A stone is thrown horizontally with an initial speed of 8 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 4.3 s. what is the height of the cliff?

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  1. 1 July, 09:05
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    90.6m

    Step-by-step explanation:

    Initial speed = 8m/s

    Time = 4.3 s

    In studying the motion of an object, we often make use of kinematic equation

    s = ut + 1/2at^2

    v^2 = u^2 + 2as

    a = (v - u) / t

    v = final velocity

    u = initial velocity

    s = distance or height

    t = time

    a = acceleration due to gravity

    Given time to hit the ground as 4.3s, we calculate the height by using

    s = ut + 1/2at^2

    Since the stone is moving against gravity, a is negative

    u = 0

    s = 0 (4.3) + 1/2 (-9.8) (4.3) ^2

    s = - 90.6m

    The height of the cliff is 90.6m
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