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9 October, 18:22

A stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5 m/s. how long is stone in the air?

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  1. 9 October, 18:23
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    There are 2 cases here:

    a. The stone travels to another certain height until its velocity becomes 0.

    i. e V² = u² - 2gh (given u = 19.5m/s)

    When it reaches maximum height V = 0

    so, h = u²/2g = (19.5) ² / (2*9.8) = > h = 19.4m

    b. Now the body starts to fall freely under gravity, so:

    h = 1/2*gt²

    We know maximum height reached by the body = height of building + height traveled after throw in (a)

    or., t = √ (2h/g)

    or, t = √2 * (19.4+59.4) / 9.8 = 4 second
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