Kinematics 2D: Horizontal Velocity of Spiked Volleyball

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The problem involves calculating the horizontal component of a volleyball's velocity after it is spiked at an initial velocity of 15 m/s at a 55-degree angle below the horizontal. The formula used for this calculation is horizontal velocity = velocity * cos(angle). The calculated horizontal velocity is 8.6 m/s. The poster expresses uncertainty about the simplicity of the solution, questioning if it could be correct. Overall, the calculation appears accurate based on the provided information.
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Homework Statement


A volleyball is spiked so that it has an initial velocity of 15m/s directed downward at an angle of 55degrees below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?

Homework Equations


horizontal velocity = velocity * cos(angle)


The Attempt at a Solution


15m/s * cos(55) = 8.6m/s
It seems like I'm missing something.
 
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Looks good to me.
 
Really? I spent a long time on this thinking it was too easy to be correct...haha.
Thank you!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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