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27 March, 10:27

A long jumper jumps at a 20 degree angle and attains a maximum altitude of 0.6 m

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  1. 27 March, 12:09
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    Refer to the diagram below.

    Assume g = 9.8 m/s² and ignore air resistance.

    Let V = the launch velocity.

    The horizontal component of the launch velocity is

    Vx = V cos (20°) = 0.9397V m/s

    The vertical component of the launch velocity is

    Vy = V sin (20°) = 0.342V m/s

    Let t = the time to attain the maximum height of 0.6 m.

    At maximum height, the vertical velocity is zero, therefore

    Vy - 9.8 t = 0

    Vy = 9.8t (1)

    Also,

    Vy*t - (1/2) * 9.8*t² = 0.6 (2)

    Substitute (1) into (2).

    (9.8t) * t - 0.5*9.8*t² = 0.6

    4.9t² = 0.6

    t = 0.35 s

    Therefore,

    Vy = 9.8*0.35 = 3.43 m/s

    V = Vy/0.342 = 10.0292 m/s

    Vx = 0.9397V = 9.4244 m/s

    The total time of travel is 2t = 2*0.35 = 0.7 s.

    The horizontal travel is

    d = 9.4244 * 0.7 = 6.597 m

    Answers:

    The time for the jump is 0.7 s.

    The length of the jump is 6.6 m (nearest tenth)
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