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A body is projected vertically upwards at time $t=0$ and it is seen at a height $H$ at time $t_1$ and $t_2$ second during its flight. The maximum height attained is ( $g$ is acceleration due to gravity)
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Verified Answer
The correct answer is:
$\frac{g\left(t_2-t_1\right)^2}{8}$
Let time taken by the body to fal from point $C$ to $B$ is $t^{\prime}$.
Then $t_1+2 t^{\prime}=t_2$
$t^{\prime}=\left(\frac{t_2-t_1}{2}\right) \ldots(\mathrm{i})$
Total time taken to reach point $C$
$\begin{aligned}
T & =t_1+t^{\prime} \\
& =t_1+\frac{t_2-t_1}{2} \\
& =\frac{2 t_1+t_2-t_1}{2} \\
& =\left(\frac{t_1+t_2}{2}\right)
\end{aligned}$
Maximum height attained
$\begin{aligned}
H_{\max } & =\frac{1}{2} g(T)^2 \\
& =\frac{1}{2} g\left(\frac{t_1+t_2}{2}\right)^2 \\
& =\frac{1}{2} g \cdot \frac{\left(t_1+t_2\right)^2}{4} \\
\Rightarrow \quad H_{\max } & =\frac{1}{8} g \cdot\left(t_1+t_2\right)^2 \mathrm{~m}
\end{aligned}$
Then $t_1+2 t^{\prime}=t_2$
$t^{\prime}=\left(\frac{t_2-t_1}{2}\right) \ldots(\mathrm{i})$
Total time taken to reach point $C$
$\begin{aligned}
T & =t_1+t^{\prime} \\
& =t_1+\frac{t_2-t_1}{2} \\
& =\frac{2 t_1+t_2-t_1}{2} \\
& =\left(\frac{t_1+t_2}{2}\right)
\end{aligned}$
Maximum height attained
$\begin{aligned}
H_{\max } & =\frac{1}{2} g(T)^2 \\
& =\frac{1}{2} g\left(\frac{t_1+t_2}{2}\right)^2 \\
& =\frac{1}{2} g \cdot \frac{\left(t_1+t_2\right)^2}{4} \\
\Rightarrow \quad H_{\max } & =\frac{1}{8} g \cdot\left(t_1+t_2\right)^2 \mathrm{~m}
\end{aligned}$
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