An object is dropped from a height of 22 m. At what height will its kinetic energy and its potential energy be equal?

5 Answers

  • ?
    6 days ago

    at a stationary height of 22 m , the object has only potential energy and no kinetic energy , so initially its total energy = potential energy only , now while falling down it looses the height and simultaneously its potential energy also goes on reducing but at the same time equal amount of energy is getting transformed in to its kinetic energy which is the principle of conservation of energy, so when it reaches half the distance of 22 m , that is 11 meter , it will attain equal amount of potential energy and kinetic energy

  • sojsail
    6 days ago

    The total energy is constant through the entire fall. Total energy, TE, is given by

    TE = gravitational potential energy + kinetic energy

    At the moment of drop, all the TE is in the form of gravitational potential energy. At a height of 11 m, gravitational potential energy is 1/2 of what it was initially. (Gravitational potential energy = m*g*h.)

    Since TE must still have the same value (because of the principle of conservation of energy), KE must also be 1/2 of the initial value.

  • donpat
    6 days ago

    Half of the initial potential energy will have been converted to kinetic energy at a

    height of 11 m and the potential energy and kinetic energy will equal one half of

    the initial potential energy.

    Answer : 11 m <——–

    PE2 = KE2 = ( 1/2 ) ( PE1 )

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  • Robert B.
    6 days ago

    I’m not sue of this answer and I’m probably wrong but I would guess 22m.

  • ?
    6 days ago


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