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23 June, 08:36

A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after reaching a height of 3.0 m. (a) What is the velocity of the football when it first reaches a height of 3.0 m above the level ground? (b) What horizontal distance has the ball travelled when it first reaches a height of 3.0 m above ground?

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  1. 23 June, 11:24
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    Refer to the diagram shown below.

    Neglect air resistance.

    The horizontal component of the launch velocity is

    (20 m/s) * cos (37°) = 15.973 m/s

    The vertical component of the launch velocity is

    (20 m/s) * sin (37°) = 12.036 m/s

    The acceleration due to gravity is g = 9.8 m/s².

    The time, t s, for the ball to reach a height of 3 m is given by

    (12.036 m/s) * (t s) - (1/2) * (9.8 m/s²) * (t s) ² = (3 m)

    12.036t - 4.9t² - 3 = 0

    2.4543t - t² - 0.6122 = 0

    t² - 2.4563t + 0.6122 = 0

    Solve with the quadratic formula.

    t = (1/2) [2.4563 + / - √ (6.0334 - 2.4488) ]

    t = 2.1748 or 0.2815 s

    The ball reaches a height of 3 m twice.

    The first time it reaches 3 m height is 0.2815 s.

    Part a.

    The vertical velocity when t = 0.2815 s is

    Vy = 12.036 - 9.8*0.2815

    = 9.2773 m/s

    The horizontal component of velocity is Vx = 15.973 m/s

    The resultant velocity is

    √ (9.2773² + 15.973²) = 18.47 m/s

    Answer:

    The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)

    Part b.

    The horizontal distance traveled is

    d = (15.973 m/s) * (0.2815 s) = 4.4964 m

    Answer:

    The horizontal distance traveled is 4.5 m (nearest tenth)
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