Question

A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must rise at least 0.330m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket? Please show wooooorkkkkk =)

Answer 1

We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. 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 (2 sig figs)

Answer 2

Answer:

48.6 m/s

Explanation:

height, h = 0.33 m

angle of projection, θ = 3°

Let the velocity of projection is u.

Use the formula for the maximum height raised

$H=\frac{u^{2}Sin^{2}\theta }{2g}$

$0.33=\frac{u^{2}Sin^{2}3 }{2\times 9.8}$

u = 48.6 m/s