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Conservation of mechanical energy

Question

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)

Moderate

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
$∴ mgh = \frac{490}{2} \Rightarrow 2 \times 9 .8 \times h = \frac{490}{2} \Rightarrow h = 12 .5 m .$

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