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25 May, 08:59

A basketball is thrown up into the air. It is released with an initial velocity of 8.5 m/s. How long does it take to get to the top of its motion?

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Answers (2)
  1. 25 May, 09:39
    0
    It takes 0.867 seconds to get to the top of its motion

    Explanation:

    We have equation of motion v = u + at

    Initial velocity, u = 8.5 m/s

    Final velocity, v = 0 m/s - At maximum height

    Time, t = ?

    Acceleration, a = - 9.81 m/s²

    Substituting

    v = u + at

    0 = 8.5 + - 9.81 x t

    t = 0.867 s

    It takes 0.867 seconds to get to the top of its motion
  2. 25 May, 12:15
    0
    0.87 s

    Explanation:

    initial velocity, u = 8.5 m/s

    Let it takes time t to reach to maximum height. At maximum height the velocity is zero, so, v = 0

    Use first equation of motion

    v = u - gt

    where, g be the acceleration due to gravity

    0 = 8.5 - 9.8 t

    t = 0.87 s

    Thus, the time taken to reach at top is 0.87 s.
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