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 ><

 

[baba89]

To calculate the impulse delivered to the ball by the floor, we need to consider the change in momentum of the ball. The equation for impulse is I = Δp = mΔv, where m is the mass of the ball and Δv is the change in velocity. In this case, we know the mass of the ball (0.60 kg) and the initial and final velocities (6.5 m/s). To account for the angle, we can use trigonometry to find the vertical component of the velocity, which is the change in velocity in the y-direction. This can be calculated as Δvy = v*sin(theta) = (6.5 m/s)*sin(58°) = 5.7 m/s. Therefore, the impulse delivered to the ball by the floor is I = mΔvy = (0.60 kg)*(5.7 m/s) = 3.42 kg*m/s. This means that the floor exerted a force on the ball for a certain amount of time (Δt) to change its momentum by 3.42 kg*m/s. I hope this helps!