Question

a stone is dropped from a height of 100 m. At what height is its kinetic energy is 2/5 of it's potential energy

Answer 1

Let at a particular distance the height is x and the stone is dropped from the height h.

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KE

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgx

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx=

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h The new kinetic energy is given as,

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h The new kinetic energy is given as,KE

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h The new kinetic energy is given as,KE n

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h The new kinetic energy is given as,KE n

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h The new kinetic energy is given as,KE n =4mgx

Let at a particular distance the height is x and the stone is dropped from the height h.The potential energy is given as,PE=9×KEmg(h−x)=9×mgxx= 10h The new kinetic energy is given as,KE n =4mgxKE =

n

n

n =4mg×

n =4mg× 10

n =4mg× 10h

n =4mg× 10h

n =4mg× 10h

n =4mg× 10h KE =

n

n

n =

n = 5

n = 52

n = 52

n = 52 mgh

n = 52 mghThe potential energy is given as,

n = 52 mghThe potential energy is given as,PE

n = 52 mghThe potential energy is given as,PE n

n = 52 mghThe potential energy is given as,PE n

n = 52 mghThe potential energy is given as,PE n =

n = 52 mghThe potential energy is given as,PE n = 5

n = 52 mghThe potential energy is given as,PE n = 53

n = 52 mghThe potential energy is given as,PE n = 53

n = 52 mghThe potential energy is given as,PE n = 53 mgh

n = 52 mghThe potential energy is given as,PE n = 53 mghThe required ratio is given as,

n = 52 mghThe potential energy is given as,PE n = 53 mghThe required ratio is given as,r=

n = 52 mghThe potential energy is given as,PE n = 53 mghThe required ratio is given as,r= 3

n = 52 mghThe potential energy is given as,PE n = 53 mghThe required ratio is given as,r= 32

n = 52 mghThe potential energy is given as,PE n = 53 mghThe required ratio is given as,r= 32

Answer 2

see the attach pic.....