physics

The ball is thrown with an initial speed of 3.7 m/s at an angle of 15.4° below the horizontal. It is released 0.83 m above the floor. What horizontal distance does the ball cover before bouncing?

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  1. (horizontal velocity component) x (time to fall) =
    (3.7 cos 15.4)* T

    Get T by solving the equation
    3.7 sin 15.4*T + (g/2) T^2 = 0.83

    Take the positive root.

    posted by drwls

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