What is the Height at Which a Falling Ball Reaches a Velocity of 5.2 m/s?

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SUMMARY

The discussion focuses on calculating the height at which a falling ball reaches a velocity of 5.2 m/s, starting from an initial height of 5 meters. Utilizing the conservation of energy theorem, the equation m*g*hi - (1/2) * m * vf² = mghf is derived, where m represents mass, g is gravitational acceleration, hi is the initial height, vf is the final velocity, and hf is the final height. The mass cancels out in the equation, allowing for a solution without needing its value. The final height can be determined through algebraic manipulation of the energy equations.

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Homework Statement


A ball falls from rest from an initial height of 5m above the ground. At what height above ground level will the ball be moving at 5.2 m/s?

Homework Equations


gravitational energy = mgh
kenetic energy=(1/2)mv^2

The Attempt at a Solution


I tried solving for both types of energy but I do not have mass so I am not sure how to solve. Please help!
 
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What you want to do is try using the conservation of energy theorem:
Ki + Pi = Kf + Pf
It starts from rest so Ki = 0, we want to find hf, which is in Pf, so solve for that

In the end your equations should work out to be
m*g*hi - (1/2) * m *vf2 = mghf
Use algebra to solve for hf. You can see that all the masses cancel, so you don't need them, and the velocity you have is vf! Let me know if it works.
 
Thank you! It worked!
 

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