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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)
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Verified Answer
The correct answer is:
$\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2$
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
$$
\begin{aligned}
\mathrm{h} & =\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right) \mathrm{t}_1-\frac{1}{2} \mathrm{gt}_1^2 \\
\mathrm{~h} & =\frac{1}{2} \mathrm{gt}_1\left(\mathrm{t}_1+\mathrm{t}_2-\mathrm{t}_1\right) \\
\therefore \quad \mathrm{h} & =\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2
\end{aligned}
$$
$$
\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2
$$
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
$$
\begin{aligned}
\mathrm{h} & =\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right) \mathrm{t}_1-\frac{1}{2} \mathrm{gt}_1^2 \\
\mathrm{~h} & =\frac{1}{2} \mathrm{gt}_1\left(\mathrm{t}_1+\mathrm{t}_2-\mathrm{t}_1\right) \\
\therefore \quad \mathrm{h} & =\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2
\end{aligned}
$$
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