What height is KE equal to PE for a 4kg ball thrown at 40m/s?

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Homework Help Overview

The discussion revolves around determining the height at which the kinetic energy (KE) of a 4 kg ball thrown upwards at 40 m/s is equal to its potential energy (PE). The problem involves concepts from energy conservation in physics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic and potential energy, referencing the conservation of energy principle. There are questions about relevant equations and initial conditions, such as the starting height of the ball.

Discussion Status

Some participants have suggested using the conservation of energy equation to find the height where KE equals PE. There is an ongoing exploration of the problem, with attempts to clarify initial conditions and the application of energy equations. No consensus has been reached on the final answer.

Contextual Notes

Participants are working under the assumption that the ball is thrown from a height of 0 m. There is mention of specific energy values, but the discussion remains open-ended regarding the final solution.

zafer
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when a ball with 4 kg in thrown upwards at a velocity of 40m/s, at what height is the KE equal to PE?
 
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Do you have any relevant equations yet?
Also, I assume it is being thrown from a y(0)= 0m?
 
It is like KEi+PEi=KEf+PEf and I need to find at which height they are equal

yes it is thrown from y=0m
 
[tex]\frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mgh_2[/tex]
As the equation shows, they are equal at the height that takes half of the initial kinetic energy.
 
can you please solve it xcvxcvvc?
 
zafer said:
can you please solve it xcvxcvvc?

Solve what? You don't use the entire equation I wrote. You solve the equation I defined with my words.
 
Have you tried anything yet?
 
yes it should be h=40 m to be equal.
Because the KEi=3200 J and it should take the half of it to be equal,which means 1600J
 

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