A 1.85 m tall basketball player attempts a goal 9.3 m from a basket (3.05 m high). If he shoots the ball at a 45° angle, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard?

Answer 1

hf=hi+vi'*t+1/2 g t^2
where vi' is the vertical component of speed initially (Vsin45)

In the horizontal
d=VCos45*t
now t is the same in each equation, form the second, t=9.3*1.414/V; put that in the first equation for t..

3.05=1.85+V*.707*(9.3*1.414/V)+4.9(9.3*1.414/V)^2

multiply through both sides by V^2, then you have a quadratic, solve using the qudaratic equation.