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5 June, 21:49

A handball is hot toward a wall with a velocity of 13.7 m/s in the forward direction. It returns with a velocity of 11.5 m/s in the backward direction. If the time interval during which the ball is accelerated is 0.021s, what is the handballs average acceleration?

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  1. 5 June, 23:19
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    1200 m/s2

    Explanation:

    a = (v-u) / t

    where a = acceleration

    v = final velocity

    u = initial velocity

    t = time taken

    As velocity is a vector it is having a direction. So as the ball bumps out in the opposite direction which it was thrown, the sign of the velocities must be different from each other.

    Consider u = - 13.7 m/s

    then v = + 11.5 m/s

    t = 0.021 s

    Applying in the above equation,

    a = (11.5 - (-13.7)) / 0.021 = 25.2/0.021 = 1200 m/s2
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