Question

The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 8.00 m/s, releasing it at a height of 2.48 m above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error.
° above the horizontal

Answer 1

Ignoring ball and hoop diameter, just solve

y(x) = -gsec^2(θ)/2v^2 x^2 + tanθ x + h
when y(4.57) = 3.05

-9.8 sec^2θ/(2*8^2) * 4.57^2 + tanθ (4.57) + 2.48 = 3.05

-1.6 (1+tan^2θ) + 4.57 tanθ - 0.57 = 0
1.6 tan^2θ - 4.57 tanθ + 2.17 = 0

tanθ = .602 or 2.255
θ = 31° or 66°

Answer 2

Steve,
Do you mind explaining where you derived the y(x) equation?
It looks like something I had before I looked for help on this problem and I am just trying to clarify just in case I see this problem again.
I did use this help to get the problem answered correctly and am just curious now.
Any feedback would be great