Question

A body is falling freely what will be ratio of kinetic energy and potential energy at mid point

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Answer 1

Potential Energy:-

$\boxed{\sf E_P=mgh}$

• EP Denotes to potential energy
• m denotes to mass of body
• g=acceleration due to gravity.
• h=height

Kinetic energy:-

$\boxed{\sf E_K=\dfrac{1}{2}mv^2}$

• Ek=Kinetic energu
• m denotes to mass of body
• v denotes to velocity of body

Ratio:-

$\\ \sf\longmapsto \dfrac{E_k}{E_P}=\dfrac{\dfrac{1}{2}mv^2}{mgh}$

$\\ \sf\longmapsto \dfrac{E_K}{E_P}=\dfrac{v^2gh}{2}$

$\\ \sf\longmapsto E_K:E_P=v^2gh:2$