With 2.0 seconds left, a basketball player 15.0m away from the 3.0m high hoop throws the ball. Determine the direction and speed of the ball upon its release at a height of 2.0m above the ground, such that it enters the hoop as the referee blows the final whistle.
Answer: 13m/s at 54° above the horizontal
Anyone can explain how to get the answer?
Thanks in advance.
Hi,
Consider horizontal motion, then
v sin theta = 15/2 -- (1),
since horizontal speed is constant, the player is 15.0m away, and the ball must enter the hoop within 2.0s.
Consider vertical motion, then
(v sin theta)t - 1/2 gt^2 = 1
(v sin theta)(2) - 1/2 (10)(2^2) = 1
v sin theta = 21/2 -- (2),
using the result s = ut + 1/2 at^2 and taking upward direction as positive.
To find theta, consider (1) / (2); to find v, consider (1)^2 + (2)^2.
Thanks!
Cheers,
Wen Shih