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A body is thrown vertically upwards from A, the top of the tower, reaches the ground in time $t_{1}$. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time $\mathrm{t}_{2}$. If it is allowed to fall freely from $\mathrm{A},$ then the time it takes to reach the ground is given by
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Verified Answer
The correct answer is:
$\quad \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}$
Let the body is projected vertically upwards from A with a speed $u_{0}$ Using equation, $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$
For case $(1)-\mathrm{h}=\mathrm{u}_{0} \mathrm{t}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2} \quad \ldots(1)$
For case (2) $-\mathrm{h}=\mathrm{u}_{0} \mathrm{t}_{2}-\frac{1}{2} \mathrm{gt}_{1}^{2}$
Subtracting eq (2) from (1), we get
$$
\begin{array}{l}
0=\mathrm{u}_{0}\left(\mathrm{t}_{2}+\mathrm{t}_{1}\right)+\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{2}^{2}-\mathrm{t}_{1}^{2}\right) \\
\Rightarrow \mathrm{u}_{0}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)
\end{array}
$$
Putting the value of $\mathrm{u}_{0}$ in eq (2), we get
$$
-\mathrm{h}=-\left(\frac{1}{2}\right) \mathrm{g}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right) \mathrm{t}_{2}-\left(\frac{1}{2}\right) \mathrm{gt}_{2}^{2}
$$
$$
\Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1} \mathrm{t}_{2}\right)
$$
For case $3, \mathrm{u}_{0}=0, \mathrm{t}=?$
$$
-\mathrm{h}=0 \times \mathrm{t}-\left(\frac{1}{2}\right) \mathrm{gt}^{2}
$$
$\mathrm{h}=\left(\frac{1}{2}\right) \mathrm{gt}^{2}$
Comparing eq. (4) and $(5),$ we get
$$
\frac{1}{2} \mathrm{gt}^{2}=\frac{1}{2} \mathrm{gt}_{1} \mathrm{t}_{2} \quad \therefore \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}
$$
For case $(1)-\mathrm{h}=\mathrm{u}_{0} \mathrm{t}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2} \quad \ldots(1)$
For case (2) $-\mathrm{h}=\mathrm{u}_{0} \mathrm{t}_{2}-\frac{1}{2} \mathrm{gt}_{1}^{2}$
Subtracting eq (2) from (1), we get
$$
\begin{array}{l}
0=\mathrm{u}_{0}\left(\mathrm{t}_{2}+\mathrm{t}_{1}\right)+\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{2}^{2}-\mathrm{t}_{1}^{2}\right) \\
\Rightarrow \mathrm{u}_{0}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)
\end{array}
$$
Putting the value of $\mathrm{u}_{0}$ in eq (2), we get
$$
-\mathrm{h}=-\left(\frac{1}{2}\right) \mathrm{g}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right) \mathrm{t}_{2}-\left(\frac{1}{2}\right) \mathrm{gt}_{2}^{2}
$$
$$
\Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1} \mathrm{t}_{2}\right)
$$
For case $3, \mathrm{u}_{0}=0, \mathrm{t}=?$
$$
-\mathrm{h}=0 \times \mathrm{t}-\left(\frac{1}{2}\right) \mathrm{gt}^{2}
$$
$\mathrm{h}=\left(\frac{1}{2}\right) \mathrm{gt}^{2}$
Comparing eq. (4) and $(5),$ we get
$$
\frac{1}{2} \mathrm{gt}^{2}=\frac{1}{2} \mathrm{gt}_{1} \mathrm{t}_{2} \quad \therefore \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}
$$
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