What is the impulse delivered by the floor in a 2D bounce pass?

[xthursday]

Homework Statement

To make a bounce pass, a player throws a 0.60 kg basketball toward the floor. The ball hits the floor with a speed of 6.5 m/s at an angle of 58° from the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?

Homework Equations

I = F*Δt = Δp = Δ(mv)

The Attempt at a Solution

P= (.60kg)*(6.5m/s)= 3.9 kg*m/s

I'm not sure what to do about the angle, theta. None of the example problems I've seen have used it. I tried multiplying the momentum by cos(theta) but I'm not getting the right answer. I would greatly appreciate it if someone could help me out with the equation for this and explain it =D

 

[xthursday]

I'm not sure if this is the right equation, but it seems like something along the lines of what I need:
m*v*cos(Ѳ) + m*v*sin(Ѳ)

The answer I got was 4.133 kg*m/s. This answer is within the correct range, but the input box says it cannot evaluate my answer ><