physics
posted by !!!! on .
The ball is thrown with an initial speed of 3.7 m/s at an angle of 15.4° below the horizontal. It is released 0.83 m above the floor. What horizontal distance does the ball cover before bouncing?

(horizontal velocity component) x (time to fall) =
(3.7 cos 15.4)* T
Get T by solving the equation
3.7 sin 15.4*T + (g/2) T^2 = 0.83
Take the positive root.