A small body is thrown vertically up, which reaches a maximum height of H, and whose total time of flight is T. Its velocity at a height ‘h’ will be                          

Given that

A body is thrown up vertically

Initial velocity = u

Acceleration $a=-\text{g}$

Height ascended = h

Final velocity = v

We get

${\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=-2\text{gh}$

$\Rightarrow {\text{v}}^{\text{2}}={\text{u}}^{\text{2}}-2\text{gh}\to \left(1\right)$

Maximum height $\text{H}=\frac{{\text{u}}^{\text{2}}}{\text{2g}}$

Time of flight $\text{T}=\frac{\text{2u}}{\text{g}}$

${\text{u}}^{\text{2}}=2\text{gH}$

Now equation (1) will become

${\text{v}}^{\text{2}}=2\text{gH}-2\text{gh}$

$\Rightarrow {\text{v}}^{\text{2}}=2\text{g}\left(\text{H}-\text{h}\right)$

$\text{H}=\frac{{\text{u}}^{\text{2}}}{\text{2g}}\text{T}=\frac{\text{2u}}{\text{g}}$

$\text{H}=\frac{{\text{u}}^{\text{2}}g}{{\text{2g}}^2}\frac{\text{u}}{\text{g}}=\frac{\text{T}}{\text{2}}$

$\text{H}=\frac{\text{g}}{\text{2}}{\left(\frac{\text{u}}{\text{g}}\right)}^2$

$\text{H}=\frac{\text{g}}{\text{2}}\left(\frac{{\text{T}}^2}{4}\right)$

$\text{g}=\frac{\text{8H}}{{\text{T}}^{\text{2}}}$

${\text{v}}^2=2\times \frac{\text{8H}}{{\text{T}}^{\text{2}}}\left(\text{H}-\text{h}\right)$               

${\text{v}}^2=\frac{\text{16}}{{\text{T}}^{\text{2}}}\left({\text{H}}^2-\text{Hh}\right)$

$\text{v}=\frac{4}{\text{T}}\sqrt{{\text{H}}^2-\text{Hh}}$