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4 March, 06:04

A ball thrown straight upwards with an initial velocity of 15 m/s. Calculate how long the ball will remain in the air and the maximum height the ball reaches?

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  1. 4 March, 06:08
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    The ball remain in air for 3.06 seconds.

    Maximum height reached = 22.94 m

    Explanation:

    We have equation of motion v = u + at

    At maximum height, final velocity, v = 0 m/s

    Initial velocity = 15 m/s

    acceleration = - 9.81 m/s²

    Substituting

    0 = 15 - 9.81 t

    t = 1.53 s

    Time of flight = 2 x 1.53 = 3.06 s

    The ball remain in air for 3.06 seconds.

    We also have equation of motion v² = u² + 2as

    At maximum height, final velocity, v = 0 m/s

    Initial velocity = 15 m/s

    acceleration = - 9.81 m/s²

    Substituting

    0² = 15² - 2 x 9.81 x s

    s = 22.94 m

    Maximum height reached = 22.94 m
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