A ball is projected vertically upwards from ground. It reaches a height ' $h$ ' in time $t_1$, continues its…
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)
$\frac{1}{2} \frac{\mathrm{gt}_1}{\mathrm{t}_2}$
$\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2$
$\mathrm{g} \mathrm{t}_1 \mathrm{t}_2$
$2 \mathrm{gt}_1 \mathrm{t}_2$
Solution
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
$\begin{aligned}
\mathrm{h} & =\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right) \mathrm{t}_1-\frac{1}{2} \mathrm{gt}_1^2 \\
\mathrm{~h} & =\frac{1}{2} \mathrm{gt}_1\left(\mathrm{t}_1+\mathrm{t}_2-\mathrm{t}_1\right) \\
\therefore \quad \mathrm{h} & =\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2
\end{aligned}$