May I first of all prelude this by saying that Physics has been a huge challenge for me. So, baby steps please. 1. The problem statement, all variables and given/known data 2.05-m-tall basketball player takes a shot when he is 6.02 m from the basket. If the launch angle is 25 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. 2. Relevant equations This is an area where I struggle. If I were given the formule, I, of course, could plug in the necessary information. I think... Vo=0 Theta=25 degrees Yo=2.05m above ground. Y1=3.05m above ground Ynet=1m (Y1-Yo) Xo=0 X1=6.02m t=? (sqrt 2y/g) (positive gravity because the ball is going up...right?):uhh: v=? g=9.8m/s2 3. The attempt at a solution X=VXoT=Vo COS 25 degrees T = Vo(.99)T Y=VYoT-1/2 g(t)sqrd=Vo SIN 25 degrees = Vo(-.13)T - 1/2g(t)sqrd X=Vo.99T Y=Vo-.13T-1/2(9.8)(t)sqrd = Vo-.13T-4.9(t)sqrd This is as far as I can take it...obviously it can be taken further, but I'm not sure where or how to take it from here.