Impulse basketball homework Question

[creative_wind]

Hello,
I am (obviously) new to the forum. I am in an introductory-level physics course in university and although I had taken a year of physics in high school, these easy intro questions on my homework keep snagging me. This is online homework and I get 5 attempts at the correct solutions. This is what I have:

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 64° 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=Favg(T2-T1)=P=m*v

Where P=momentum

The Attempt at a Solution

I=m*v
m=0.6kg
v=6.5*cos(64)=2.85m/s
I=0.6kg*2.85m/s=1.71kg*m/s <--this was wrong

I=m*v
m=0.6*cos(64)=0.26kg
v=6.5*cos(64)=2.85m/s
I=0.26kg*2.85m/s=0.74kg*m/s <--this was wrong

I figured the velocity had to be in the y-component since it is vertically upward from the floor. I'm really not sure though...

Thanks :)

 

[physixguru]

this should help ya:

Impulse is the integral of force over time, it is measured in Newton-seconds. For instance a force of one Newton applied over one second will change the momentum, a force of two Newton's applied over half of a second will have a similar effect.

For rigid body collisions we take this to its limit and apply an infinite force over an infinitesimally small time. This impulse is equal to the change in momentum of the colliding objects. Because we are talking about forces here, Newton's third law applies, and the impulse on the colliding objects will be equal and opposite.

impulse = m(vf- vi)