Question

A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket. If he shoots the ball at 40 degree angle above the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? the basket height is 3.05m... please someone help me ..

Answer 1

help me.. :(

Answer 2

"Since there are decimal dimensions, we assume that the ball has negligable size, so that if it is at 3.05 m height when it has moved 10 m from the player (it has a net height with respect to the player of 1.05m), it will go through the hoop.

There are two components of velocity of interest, and call it V.
(1) the horizontal component. This is V cos40 and continues until the ball moves 10 m. Call this time T. Then 10= VT cos 40
(2) the vertical component. This is V sin40. The upwards velocity is counteracted by gravity. Since we want T to occur when the ball is 1.05 m above the height of throw,
1.05 = -4.8 T^2 + VT sin40.

You have to solve this quadradic. You will get 2 values of T. With each you get a value of V. Only one pair will be realistic; the one that works with the horizontal component equation."

source: http://answers.yahoo.com/question/index?qid=20080909110910AA90GA8