Question

A body of mass 2 kg is thrown up vertically with a KE of 490 J. If the acceleration due to gravity is $9.8m{s}^{-2}$ the height at which the KE of the body becomes half of the original value is:

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

Answer 1

The correct option is C 12.5 m

By energy conservation, the potential energy at height $h=$ kinetic energy at that height

So, $mgh=\frac{1}{2}\times 490$

or $h=\frac{490}{2mg}=\frac{490}{2\times 2\times 9.8}=12.5$ m