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17 July, 10:05

A rock is thrown upward with an initial velocity of 9.6m/s to what hight does it rise and how long does it take to reach that high

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  1. 17 July, 13:33
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    Given:

    u (initial velocity) : 9.6 m/s

    at the maximum height final velocity = 0

    The acceleration acting on the body is gravity as it is free falling object

    =-9.8m/s^2

    Now we know that

    v^2-u^2 = 2as

    Where v is the final velocity measured in m/s

    u is the initial velocity which is measured in m/s

    a is the acceleration measured in m/s^2

    s is the distance traveled by the rock in this case it is considered as the height

    Substituting these values we get

    0-9.6 = 2 x - 9.8 x s

    s = 0.49 m

    Consider t as the time taken for the rock to travel

    v=u+at

    0=9.6 - 9.8t

    t=0.98sec
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