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
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.
4 answers
is the answer 11.8 m/s ?
yes
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