Physics homework help???

Question

A basketball player shoots the ball toward a basket that is 8.00 m from him. The ball leaves his hands from a heaight of 2.00 m above the floor and at an angle of 40.0 degrees with the horizontal. The hoop of the basket is 3.05 m above the floor. What must be the initial speed of the ball so that it goes through the hoop without hitting the backboard? I know what the answer is but I don't know how to set up the equation.

Answer 1

Draw the projectile motion

For horizontal motion:

x = 0+(V_x)₀ t

x = V₀ cos 40 t

8 = V₀ cos 40 t

t = 8/ V₀ cos 40 = 10.44/V₀ 

For vertical motion:

y = y_{0 +} (V_y)₀ t - (1/2)gt²

3.05 = 2 + (V₀ sin 40)(10.44/V₀) - (1/2) (9.81)(10.44/V₀)²

1.05 =6.7107 - 534.614(1/V₀)²

534.614(1/V₀)² = 5.661

5.661(V₀)² = 534.614

(V₀)²= 94.44

V₀ = 9.72 m/s