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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4

Answers (1)

best_answer

As we discussed in

Kinetic energy -

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

- wherein

m\rightarrow mass

v\rightarrow velocity

kinetic Energy is never negative

 

 

Velocity at the  highest point =v\ cos\60^{o}=\frac{v}{2}.

K_{f}=\frac{1}{2}m(\frac{v}{2})^{2}=\frac{1}{4}(\frac{1}{2}mv^{2})

K_{f}=\frac{K}{4}

Correct option is 4


Option 1)

K/2\,

This is an incorrect option.

Option 2)

\; K\;

This is an incorrect option.

Option 3)

zero\;

This is an incorrect option.

Option 4)

K/4

This is the correct option.

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Plabita

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