Question

A particle is thrown vertically upwards. If its velocity at half of the maximum height is $10\frac{m}{s}$, then maximum height attained by it is (Take $g=10\frac{m}{s^2}$)

• A. $15m$
• B. $20m$
• C. $10m$
• D. $5m$

Answer 1

The correct option is C $10m$

Let particle be thrown with velocity $u$ and its maximum height is $H$, then $H=\frac{u^2}{2g}$.

When particle is at height $\frac{H}{2},$ then its speed is $10\frac{m}{s}$
So, from equation,
$v^2=u^2-2gh(10{)}^2=u^2-2g\left(\frac{H}{2}\right)=u^2-2g\left(\frac{u^2}{4g}\right)$
$\Rightarrow u^2=200m^2/s^2$

$\therefore$ Maximum height, $H=\frac{u^2}{2g}=\frac{200}{2\times 10}=10m$