Asked by hannah

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.

Answers

Answered by drwls
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
Answered by hannah
is the answer 11.8 m/s ?
Answered by drwls
yes
Answered by q
wwww
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