physics
posted by HELP .
In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 3.1 m/s at an angle of 15.0° below the horizontal. It is released 0.82 m above the floor. What horizontal distance does the ball cover before bouncing?
Please need help! you just have to show me how I can solve this problem

v₀(x)=v₀•cosα =3.1• cos15º=3 m/s
v₀(y)=v₀•sinα=3.1•sin15 º =0.8 m/s
h= v₀(y) •t+gt²/2.
9.8t²+0.8t0.82 =0
t= { 0.8±sqrt(0.8²+4•9.8•0.82)}/2•9.8,
The positive root is t = 0.25 s.
v(x)= v₀(x)
x= v₀(x) •t=0.75 m
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