A body of mass 2 $k g$ is thrown up vertically with $K . E .$ of 490 Joules. If the acceleration due to gravity is 9.8 $m / s^{2}$ , then the height at which the K.E. of the body becomes half its originalvalue is given by

50 m

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12.5 m

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25 m

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10 m

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Solution

The correct option is B $12.5 m$
Let $h$ is that height at which the kinetic energy of the body becomes half its original value i.e.
Half of its kinetic energy is converted into potential energy
$P E = K E / 2$

$∴ m g h = \frac{490}{2}$

$\Rightarrow 2 \times 9.8 \times h = \frac{490}{2} \Rightarrow 12.5 m .$

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A body of mass 2 kg is thrown up vertically with K.E. of 490 joule. If the acceleration due to gravity is 9.8 $m / s^{2}$ , then the height at which the K.E. of the body becomes half its original value is given by

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A body of mass 2 kg is thrown up vertically with K.E. of 490 Joules. If the acceleration due to gravity is 9.8 m/s2 , then the height at which the K.E. of the body becomes half its originalvalue is given by

The correct option is B 12.5mLet h is that height at which the kinetic energy of the body becomes half its original value i.e. Half of its kinetic energy is converted into potential energyPE=KE/2∴mgh=4902⇒2×9.8×h=4902⇒12.5m.

A body of mass 2 $k g$ is thrown up vertically with $K . E .$ of 490 Joules. If the acceleration due to gravity is 9.8 $m / s^{2}$ , then the height at which the K.E. of the body becomes half its originalvalue is given by