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A ball is thrown vertically downwards from a height of 20 m with an initial velocity \upsilon _0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity \upsilon _0 is : (Take g = 10 ms-2)

  • Option 1)

    20 ms-1     

  • Option 2)

    28 ms-1

  • Option 3)

    10 ms-1

  • Option 4)

    14ms-1

 

Answers (1)

As we learnt in

Kinetic energy -

k= frac{1}{2}mv^{2}

- wherein

m ightarrow mass

v ightarrow velocity

kinetic Energy is never negative

 When balls collide with the ground it loses its's 50% of energy

\therefore \frac{K_{f}}{K_{i}}=\frac{1}{2}

=\frac{\frac{1}{2}mv_{f}^{2}}{\frac{1}{2}mv_{i}^{2}}

\frac{v_{f}}{v_{i}}=\frac{1}{\sqrt{2}}

=> \frac{\sqrt{2gh}}{\sqrt{u^{2}+2gh}}= \frac{1}{\sqrt{2}}

4gh=u^{2}+2gh

u=20ms^{-1}


Option 1)

20 ms-1     

Correct

Option 2)

28 ms-1

Incorrect

Option 3)

10 ms-1

Incorrect

Option 4)

14ms-1

Incorrect

Posted by

Vakul

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