Asked by yenebeb
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.
Answers
Answered by
bobpursley
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
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
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