Question

A body of mass $2kg$ is thrown up vertically with kinetic energy of $490J$. If $g=9.8m/s^2$ , the height at which the kinetic energy of the body becomes half of the original values, is :

• A. $50m$
• B. $25m$
• C. $12.5m$
• D. $19.6m$

Answer 1

Answer: (C)

$12.5m$

1) The K.E becomes half when speed becomes $\frac{1}{\sqrt{2}}$ of its original value.
2) Using $2as=v^2-u^2$
$2\times (+9.8)\times s=\frac{+v^2}{2}$
Multplying both side by $m$.
$2\times (2\times 9.8\times s)=\frac{m{v}^2}{2}=490J$
$\Rightarrow s=12.5m$