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.

The horizontal component of the initial speed:

v x = v · cos 40° = v · 0.766

The vertical component of the initial speed:

v y = v · sin 40° = v · 0.6428. If the player is 10 m from the basket:

(1) 10 = v · 0.766 · t

Net high is 1.05 m with the respect to the player (3.05 m - 2 m)

1.05 = - gt²/2 + v · 0.6428 · t

(2) 1.05 = - 4.9 t² + v t · 0.6428

Now we have to solve a system of equations:

v = 13.05 / t (we will put it in the second equation)

1.05 = - 4.9 t² + 8.39

4.9 t² = 7.34

t ≈ 1.22 s

v = 13.05 : 1.22

Answer:

v = 10.07 m/s (initial speed)