A body of mass $5 k g$ is thrown vertically up with a kinetic energy of $490 J$ . The height at which the kinetic energy of the body becomes half of the original value is

$12.5 m$

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$10 m$

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$2.5 m$

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$5 m$

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Solution

The correct option is D
$5 m$

Step 1. Given data
Initial kinetic energy, $k_{i} = 490 J$
Initial potential energy, $P_{i} = 0$
Mass of the body, $m = 5 k g$
Step 2. Finding the height, $h$ at which kinetic energy is half of the original value.
According to the question,
Final kinetic energy, $k_{f} = \frac{1}{2} k_{i}$
$k_{f} = \frac{1}{2} \times 490$
$k_{f} = 245 J$
Final potential energy, $P_{f} = m g h$
Applying the energy conservation
$k_{i} + P_{i} = k_{f} + P_{f}$
$490 = 245 + 5 \times 9.8 \times h$
$h = 5 m$
Hence, the correct option is $(D)$ .

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A body of mass 5kg is thrown vertically up with a kinetic energy of 490J. The height at which the kinetic energy of the body becomes half of the original value is

A body of mass $5 k g$ is thrown vertically up with a kinetic energy of $490 J$ . The height at which the kinetic energy of the body becomes half of the original value is