posted by John Miller .
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?
hf=hi+vi'*t+1/2 g t^2
where vi' is the vertical component of speed initially (Vsin45)
In the horizontal
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..
multiply through both sides by V^2, then you have a quadratic, solve using the qudaratic equation.