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6 November, 15:58

A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40° above the horizontal. How far above or below its original level will the ball strike the opposite wall?

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  1. 6 November, 16:20
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    Ball hit the tall building 50 m away below 10.20 m its original level

    Explanation:

    Horizontal speed = 20 cos40 = 15.32 m/s

    Horizontal displacement = 50 m

    Horizontal acceleration = 0 m/s²

    Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

    t = 3.26 s

    Now we need to find how much vertical distance ball travels in 3.26 s.

    Initial vertical speed = 20 sin40 = 12.86 m/s

    Time = 3.26 s

    Vertical acceleration = - 9.81 m/s²

    Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x - 9.81 x 3.26²

    s = - 10.20 m

    So ball hit the tall building 50 m away below 10.20 m its original level
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