A body is dropped from a height $h$. Simultaneously, another body is projected vertically upwards with velocity $u$ from the ground. They meet at a height $h/2$. What is the value of $u$?

Solution

1. For the body dropped from height $h$, the displacement at the meeting point is $h/2=\frac{1}{2}g{t}^2$, which gives the time $t=\sqrt{h/g}$.
2. For the body projected upwards from the ground, the displacement at the same time $t$ is $h/2=ut-\frac{1}{2}g{t}^2$.
3. Substituting $t=\sqrt{h/g}$ into the second equation, we get $h/2=u\sqrt{h/g}-h/2$.
4. Simplifying the equation leads to $h=u\sqrt{h/g}$, which results in $u=\sqrt{gh}$.
5. Hence the answer is (A).