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10 December, 08:21

A basketball player is trying to make a half-court jump shot, and releasing the ball at the height of the basket. Assuming that the ball is launched at 51.0°, 16.0 m from the basket, what speed must the player give the ball?

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  1. 10 December, 09:31
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    The player must give a speed of 12.66 m/s to the ball.

    Explanation:

    Since, the ball is released at the same height as the basket. Therefore, this is a complete projectile motion, with following parameters:

    Horizontal Range = R = 16 m

    Launch Angle = θ = 51°

    Launch Speed = V = ?

    The formula for horizontal range of projectile is:

    R = V² Sin 2θ/g

    V² = Rg / Sin 2θ

    V² = (16 m) (9.8 m/s²) / Sin 2 (51°)

    V = √160.3 m²/s²

    V = 12.66 m/s
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