al physics mc

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  • 1 decade ago
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    Consider the horizontal component

    Horizontal speed of the ball

    =10cos30*

    =8.66 m/s

    Consider the vertical component

    Initial vertical speed of the ball

    =10sin30*

    =5 m/s

    The ball will experience a downward acceleration due to gravity.

    By the law of conservation of momentum

    Loss of P.E. = Gain of K.E.

    mgh = 1/2 mv^2

    2gh = v^2

    Vertical speed of the ball just before reached the ground, v = 14 m/s

    As the collision is perfectly elastic

    So, vertical speed of the ball just left B = 14 m/s

    By s = ut + 1/2 gt^2

    0 = 14t - 4.9t^2

    t = 2.857 s

    The ball is moving with the uniform horizontal speed

    Therefore, horizontal distance BC

    = horizontal speed X flight time

    =8.66 X 2.857

    =24.7 m

    Source(s): Myself~~~
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