Atennis player standing 12.6 m from the net hits the ball at 3.18° above the horizontal. to clear the net, the ball must rise at least 0.362 m. if the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket?

At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m.
So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY
We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION:
v² = u² + 2as
0 = u² - 2gh
u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity)
So rearranging,
velocity = (velocity in y direction only) / sin 3°
= √(2gh)/sin 3°
= (√(2 x 9.8 x 0.33)) / sin 3°
= 49 m/s at 3° to the horizontal

Ibelieve the answer you're looking for is