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8 October, 14:29

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 61 m/s^2. If he reaches the ground with a speed of 17 m/s, how long was he in the air (in seconds) ?

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  1. 8 October, 14:48
    0
    55.66 m

    Explanation:

    While falling by 50 m, initial velocity u = 0

    final velocity = v, height h = 50, acceleration g = 9.8

    v² = u² + 2gh

    = 0 + 2 x 9.8 x 50

    v = 31.3 m / s

    After that deceleration comes into effect

    In this case final velocity v = 17 m/s

    initial velocity u = 31.3 m/s

    acceleration a = - 61 m/s²

    distance traveled h = ?

    v² = u² + 2gh

    (17) ² = (31.3) ² - 2x 61xh

    h = 690.69 / 2 x 61

    = 5.66 m

    Total height during which he was in air

    = 50 + 5.66

    = 55.66 m
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