Physics help?

Question

A basketball player who is 2.00 m tall is standing on the floor L = 13.0 m from the basket, as in Figure P4.52. If he shoots the ball at a 38.0° angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m

Answer 1

Let's make some reasonable assumptions;
- The shooter is standing 13.0m from the center of the basket
- The shooter releases the ball directly above the point where he's standing and at a level at the top of his head.
Let y(zero) = 0 at 2 meters above the floor, so y(final) will be 3.05 - 2 = 1.05m

x:
(x velocity is constant)
t = D/v = D/v(x) = 13/v(x)
v(x) = v(zero)cos38

y:
v(zero y) = v(zero)sin38
y = y(zer0) + v(zero y)t + (1/2)gt^2 [remember g is negative]
y = v(zero) y)(13/v(x)) + (1/2)g(13/v(x))^2
1.05 = v(zero)sin38( 13/(v(zero)cos38)) - 4.9(13/(v(zero)cos38))^2
1.05 = 13tan38 - 1333.58/v(zero)^2
1.05 = 10.1567 - 1333.58/v(zero)^2
-9.1067 = -1333.58/v(zero)^2
v(zero)^2 = 146.439
v(zero) = 12.101m/s

Answer 2

If there were no gravity the ball would hit the shield(?) at 2+13tan(38), but because of gravity it would hit by gt^2/2 below. So gt^2/2 = 2+13tan(38) - 3.05, or t = sqrt(2/g(2 + 13tan(38) - 3.05)). During this time horizontal displacment is 13, so Vx = 13/t = 13/sqrt(2/g(2 + 13tan(38) - 3.05)). Total speed is V=Vx/cos(38).