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5 December, 09:55

A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final height of h2 = 1.38 m. Neglect air resistance. Randomized Variables m = 0.49 kg h1 = 2.25 m h2 = 1.38 m show answer No Attempt 33% Part (a) Select an expression for the impulse I that the baseball experiences when it bounces off the concrete.

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  1. 5 December, 10:23
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    Impulse = change in momentum

    mv - mu, v and u are final and initial velocity during impact at surface

    For downward motion of baseball

    v² = u² + 2gh₁

    = 2 x 9.8 x 2.25

    v = 6.64 m / s

    It becomes initial velocity during impact.

    For body going upwards

    v² = u² - 2gh₂

    u² = 2 x 9.8 x 1.38

    u = 5.2 m / s

    This becomes final velocity after impact

    change in momentum

    m (final velocity - initial velocity)

    .49 (5.2 - 6.64)

    =.7056 N. s.

    Impulse by floor in upward direction

    =.7056 N. s
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