Impulse from basket ball against ground at an angle

[okscuba45]

Homework Statement

To basketball player bounces the 0.60-kg basketball to another player by bouncing it off the court. The ball hits the court with a speed of 6.5 m/s at an angle of 50° from the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the court?

Homework Equations

I know that Impulse is the change in momentum, or the force times the change in time.

The Attempt at a Solution

I cannot figure out how to find the Impulse with simply the velocity an angles. Is there a change in momentum if all it does is change direction? If so, how do I use the angles to calculate the change?
I try doing the x/y components of the velocity but it seems to cancel out.

 

[Delphi51]

You must use xy components. There is no change in the x velocity, so no impulse. But there is a change in the y velocity - it reverses direction so you'll have something like Δv = -v - v = -2v. Of course you know the number for v, the vertical component of the 6.5 m/s.