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21 October, 03:43

A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A stop watch measures the stone's trajectory time from top of cliff to bottom to be 4.8s. What is the height of the cliff?

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  1. 21 October, 06:27
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    v = initial velocity of launch of the stone = 12 m/s

    θ = angle of the velocity from the horizontal = 30

    Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

    v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

    a = acceleration of the stone = - 9.8 m/s²

    t = time of travel = 4.8 s

    Y = vertical displacement of stone = vertical height of the cliff = ?

    using the kinematics equation

    Y = v₀ t + (0.5) a t²

    inserting the values

    Y = 6 (4.8) + (0.5) ( - 9.8) (4.8) ²

    Y = - 84.1 m

    hence the height of the cliff comes out to be 84.1 m
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