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8 March, 17:28

A shot-putter projects the shot at 42.00˚ to the horizontal from a height of 2.100 m. It lands 17.00 m away horizontally. Next, he gives it the same initial speed but changes the angle to

40.00˚. What effect does this have on the horizontal range? Watch significant figures! It matters in this

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  1. 8 March, 19:25
    0
    Let 100 m/s be the velocity of projection.

    So horizontal component

    = 100 cos42

    = 74.31 m / s

    Vertical component = - 100 sin 42. in upward direction

    66.91 m/s

    Net displacement = 2.1 downwards ( + ve)

    Using s = ut + 1/2 gt²

    2.1 = - 66.91 t +.5 x 9.8 x t²

    4.9 t² - 66.91 t - 2.1 = 0

    t = 13.685 s

    Horizontal distance covered

    = 13.685 x 74.31

    = 1016.93 m

    If angle of projction is 40°

    So horizontal component

    = 100 cos40

    = 76.60 m / s

    Vertical component = - 100 sin 42. in upward direction

    64.27 m/s

    Net displacement = 2.1 downwards ( + ve)

    Using s = ut + 1/2 gt²

    2.1 = - 76.60 t +.5 x 9.8 x t²

    4.9 t² - 76.60 t - 2.1 = 0

    t = 15.659 s

    Horizontal distance covered

    = 15.659 x 76.60

    = 1199.49 m

    So horizontal range is increased, if angle of projection is increased.
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