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23 March, 01:46

A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler's hand?

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  1. 23 March, 04:49
    0
    1.68 s

    Explanation:

    From newton's equation of motion,

    a = (v-u) / t ... Equation 1

    Making t the subject of the equation

    t = (v-u) g ... Equation 2

    Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

    Note: Taking upward to be negative and down ward to be positive,

    Given: v = 0 m/s (at the maximum height), u = 8.20 m/s, g = - 9.8 m/s²

    t = (0-8.20) / -9.8

    t = - 8.20/-9.8

    t = 0.84 s.

    But,

    T = 2t

    Where T = time taken for the bowling pin to return to the juggler's hand.

    T = 2 (0.84)

    T = 1.68 s.

    T = 1.68 s
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