A basketball player who is 2.0m tall is standing on the floor 10.0m from the basket,If he shoots the ball at a 40.0¤ angle witi the horizontal,at what initial speed must he throw so that it goes through the hoop without striking the backboard?The basket height is 3.05 m.

Answer 1

consider the vertical
hf=hi+Vsin40*t-4.9 t^2

consider the horrizontal
10=Vcos40 t
V=10/cos40t

put that in the first equation for V, then solve for t (notice it is a bit of algebra). have a pad of paper handy.

Then after you have t, V=10/tcos40