A basketball player is trying to make a half-court jump shot and releases the ball at the height of the basket. Assuming that the ball is launched at 51 degrees, 14.0 m from the basket, what speed must the player give the ball?

PLEASE HELP! I have been working on this problem for at least an hour and i still have no idea what I'm doing.

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  1. The horizontal distance that the ball travels bewtween equal elevations is

    X = 14 m = 2 (Vo^2/g)sin 51 cos 51
    = (Vo^2/g) sin 102

    (See if you can derive that by multiplying the horizontal velocity component by the time of flight)

    Vo is the "launch" velocity. Solve for it

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  2. is the answer 11.8 m/s ?

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  3. yes

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