Thank you for reporting, we will resolve it shortly
Q.
A ball is thrown vertically upward. It has a speed of $10\, m / s$ when it has reached one-half of its maximum height. How high does the ball rise? (Take, $\left.g=10 \,m / s ^{2}\right)$
The problem can be solved using third equation of motion at $A$ and $O$.
Let maximum height attained by the ball be $H$.
Third equation of motion gives $v^{2}=u^{2}-2 g h$
At $A,(10)^{2}=u^{2}-2 \times 10 \times \frac{H}{2}$
$\Rightarrow u^{2}=100+10 \,H\,\,\,\,...(i)$
At $O,(0)^{2}=u^{2}-2 \times 10 \times H$
$\Rightarrow u^{2}=20\, H\,\,\,\,...(ii)$
Thus, from Eqs. (i) and (ii),
we get $20\, H=100+10\, H$
$\Rightarrow 10 \,H=100$
$\therefore H=10 \,m$
