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 original value is given by (in m)
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answer is 12.5.
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Detailed Solution
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 will convert into potential energy∴mgh=4902 ⇒2×9.8×h=4902 ⇒h=12.5m.