A body is thrown vertically upwards at time t = 0 and it is observed at a height H at time t₁ and t₂ seconds during its flight. What will be the maximum height attained? (g is acceleration due to gravity)

When a body is projected vertically upwards it occupies the same position while going up and coming down after time t₁ and t₂ respectively.

$\begin{array}{l}\therefore \mathrm{H}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2\\ {\mathrm{gt}}^2-2\mathrm{ut}+2\mathrm{H}=0\end{array}$

Sum of roots $={\mathrm{t}}_1+{\mathrm{t}}_2=\frac{2\mathrm{u}}{\mathrm{g}}\Rightarrow \mathrm{u}$

$\begin{array}{l}=\frac{\mathrm{g}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)}{2}\\ \therefore {\mathrm{H}}_{max}=\frac{{\mathrm{u}}^2}{2\mathrm{g}}=\frac{\mathrm{g}{\left[{\mathrm{t}}_1+{\mathrm{t}}_2\right]}^2}{8}\end{array}$