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

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

When particle is at a height $H/2$, then its speed is $10m/s$

From equation $v^2=u^2-2gh$,  ${\left(10\right)}^2=u^2-2g\left(\frac{H}{2}\right)=u^2-2g\frac{u^2}{4g}$$\Rightarrow u^2=200$

 Maximum height $\therefore$= 10m