Question

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

• A. 50 m
• B. 12.5 m
• C. 25 m
• D. 10 m

Answer 1

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 will convert into potential energy

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