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!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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