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

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To determine the impulse delivered by the floor during a 2D bounce pass, the player throws a 0.60 kg basketball at 6.5 m/s and an angle of 58° from the vertical. The impulse can be calculated using the change in momentum, which involves considering both the vertical and horizontal components of the velocity. The user attempted to find the impulse by calculating momentum but struggled with incorporating the angle correctly. The calculated impulse of 4.133 kg*m/s is close but not accepted, indicating a potential error in the angle's application or the overall calculation method. Clarification on how to properly account for the angle in the impulse formula is needed for an accurate solution.
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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
 
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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 ><
 
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