i have a physics question.

Question

A basketball player who is 2.00 m tall is standing on the floor 7 m from the basket. If he shoots the ball at a 43⁰ angle with the horizontal, what initial speed must he throw the ball so that it goes through the hoop without striking the blackboard. The basket height is 3.05 m.

Answer 1

Let h = height ... then h(t) = ho+Vyo*t-1/2*g*t^2 and x(t) = xo+Vxo*t ... also Vxo=V*cos(43) & Vyo=V*sin(43) ... Let T represent the time when the ball reaches the hoop .. Find V

h(T) =3.04=2+V*sin(43)T-9.81/2*T^2

x(T) = 7 = V*cos(43)*T

Solve the above for V to get V= 9.03 m/s and T = 1.06 sec