A 3 point jump shot is released 2.2 m above the ground, 6.14 m from the basket, which is 3.06 m high. For launch angles of 30° and 60°, find the speed needed to make the basket.

Answer 1

hf=hi+vi*t -4.9t^2 in the vertical
3.06=2.2+Vsin30*t-4.9t^2 for the 30 degree shot.

then in the horizontal
5.14=Vcos30*t solve for t in terms of V
t=5.14/Vcos30

Now put that t in the first equation, solve for V. It will take some algebra, but in the end, you will get a quadratic. Use the quadratic equation.

Then do the same for the other angle. You should get the same V, think out in the math why.