A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is (Take $g=10$  m/s²) 

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}$

$\Rightarrow u^2=200$

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