A ball is projected vertically upwards from ground. It reaches a height ' \(h\) ' in time \(t_1\), continues its motion and then takes a time \(t_2\) to reach ground. The height \(h\) in terms of \(g, t_1\) and \(t_2\) is \((g=\) acceleration due to gravity)

We know,
\(
\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2
\)
The total time required for the ball to go up and reach the ground is \(t=t_1+t_2\), and the total displacement is zero.
\(
\begin{aligned}
& \therefore \quad 0=\mathrm{u}\left(\mathrm{t}_1+\mathrm{t}_2\right)+\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right)^2 \\
& \therefore \quad \mathrm{u}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right)
\end{aligned}
\)
The displacement in time \(t_1\) is
\(
\begin{aligned}
\mathrm{h} & =\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right) \mathrm{t}_1-\frac{1}{2} \mathrm{gt}_1^2 \\
\mathrm{~h} & =\frac{1}{2} \mathrm{gt}_1\left(\mathrm{t}_1+\mathrm{t}_2-\mathrm{t}_1\right) \\
\therefore \quad \mathrm{h} & =\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2
\end{aligned}
\)