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4 June, 07:39

A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the bottom of the cliff? (b) What is its speed just before hitting? And (c) what total distance did it travel? (Ans: 5.2 sec, 38.94 m/s, 84.7 m)

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  1. 4 June, 07:58
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    Answer

    given,

    vertical speed of stone, v = 12 m/s

    height of the cliff = 70 m

    a) time taken by the stone to reach at the bottom of the cliff

    We know that,

    S = u t + 1/2 a t²

    - 70 = 12 t - 0.5 x 9.8 t²

    4.9 t² - 12 t - 70 = 0

    solving the equation

    t = 5.2 s (neglecting the negative value)

    b) again using equation of motion

    v = u + a t

    v = 12 - 9.8 x 5.2

    v = - 38.96 m/s

    ignoring the negative sign

    magnitude of velocity is equal to 38.96 m/s

    c) total distance travel by the stone

    vertical distance covered by the stone

    v² = u² + 2 g h

    0 = 12² - 2 x 9.8 x h

    h = 7.34 m

    to reach the stone to the same level distance travel be doubled.

    Total distance travel by the stone

    H = h + h + 70

    H = 7.34 x 2 + 70

    H = 84.7 m.
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