A particle is projected vertically upwards and it reaches the maximum height H in time T second. The height of the particle at any time t will be

Using v = u+ gt, we have
0 = u-gT or u=gT
Further v² =u² +2g s or 0 = u² -2gH
$\therefore \mathrm{H}=\frac{{\mathrm{u}}^2}{2\mathrm{g}}=\frac{{\mathrm{g}}^2{\mathrm{T}}^2}{2\mathrm{g}}=\frac{{\mathrm{gT}}^2}{2}....\left(1\right)$
Let h be the distance travelled in time f , then
$\mathrm{h}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2=\mathrm{gtT}-\frac{1}{2}{\mathrm{gt}}^2.....\left(2\right)$
Subtracting eq. (1) from (2), we get
$\begin{array}{l}\mathrm{h}-\mathrm{H}=\mathrm{gtT}-\frac{1}{2}{\mathrm{gt}}^2-\frac{{\mathrm{gT}}^2}{2}\\ =-\frac{\mathrm{g}}{2}(\mathrm{T}-\mathrm{t}{)}^2\\ \therefore \mathrm{h}=\mathrm{H}-\frac{\mathrm{g}}{2}(\mathrm{T}-\mathrm{t}{)}^2\end{array}$