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A 2.05m--tall basketball player takes a shot when he is 6.02m from the basket. If the launch angle is 27 degrees and the ball was launched at the level of the player’s head, what must be the release speed of the ball for the player to make the shot? The basket is 3.05m above the floor.
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basket 3.05m -player 2.05m =1m
I need the velocity.
27 degrees = 3.703 +- .00703703
h = 1m 1m x 3.704 = 3 .704
I got it.
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