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19 March, 11:35

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 27 m/s^2. If he reaches the ground with a speed of 14 m/s, how long was he in the air (in seconds) ?

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  1. 19 March, 11:47
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    3.83 s

    Explanation:

    Initially for 50 m A falls with acceleration equal to g ie 9.8 m / s². In this journey his initial velocity u = 0, a = 9.8, v = ?, t = ?

    h = ut + 1 / 2 g t²

    50 =.5 x 9.8x t²

    t = 3.19 s

    v = u + gt

    = o + 9.8 x 3.19

    = 31.26 m / s

    For motion under deceleration

    initial speed u = 31.26 m/s

    Final speed v = 14 m/s

    deceleration a = - 27 m/s²

    v = u - at

    14 = 31.26 - 27 t

    t = 0.64 s

    So total time in the air

    = 3.19 +.64 = 3.83 s
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