Question on a ball dropping( conservation of energy)

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SUMMARY

A ball with a mass of 1 kg is dropped from a height of 7 meters, resulting in a potential energy (P.E.) of 70 Joules just before impact, calculated using the formula P.E. = mgh. The kinetic energy (K.E.) just before impact is also 70 Joules, as energy is conserved in free fall. To find the initial rebound velocity and kinetic energy after rebounding to a height of 4.5 meters, the same potential energy formula should be applied, adjusting for the new height.

PREREQUISITES
  • Understanding of gravitational potential energy (P.E. = mgh)
  • Knowledge of kinetic energy formula (K.E. = 1/2 mv²)
  • Basic principles of conservation of energy
  • Ability to perform calculations involving height and mass
NEXT STEPS
  • Calculate the rebound kinetic energy using P.E. = mgh for a height of 4.5 meters
  • Determine the initial rebound velocity using K.E. = 1/2 mv²
  • Explore the concept of energy loss during impact and its implications
  • Investigate real-world applications of conservation of energy in mechanics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation principles, as well as educators seeking to enhance their teaching methods in these topics.

Peter-John King
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Homework Statement



A ball of mass 1 Kg is dropped from a height of 7m and rebounds to a height of 4.5m Calculate:
its kinetic energy just before impact
the initial rebound velocity of the ball and kinetic energy. Account for loss of kinetic energy on impact



2. Homework Equations [/B]
i am not sure of all the relevant equations i used pe=mgh and k.e=1/2mv^2

The Attempt at a Solution


I found the answer to a using the formula P.E.=mgh=1x10x7=70J
K.E.=P.E.=70J
I do not know how to find the second answer
 
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Peter-John King said:
I found the answer to a using the formula P.E.=mgh=1x10x7=70J
K.E.=P.E.=70J
Good.

Peter-John King said:
I do not know how to find the second answer
Use the same equation. (A different height, of course.)
 

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