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A stone thrown down with a seed $u$ takes a time $t_{1}$ to reach the ground, while another stone, thrown upwards from the same point with the same speed, takes time $t_{2}$. The maximum height the second stone reaches from the ground is
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Verified Answer
The correct answer is:
$\mathrm{g} / 8\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$
$-\mathrm{h}=-\mathrm{Ut}_{1}+\frac{-1}{2} \mathrm{~g} \mathrm{t}_{1}{ }^{2}$
$\mathrm{~h}=\mathrm{Ut}_{1}+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}{ }^{2} \ldots .$(1)
$-\mathrm{h}=\mathrm{Ut}_{2}-\frac{1}{2} \mathrm{~g} \mathrm{t}_{2}{ }^{2} \ldots \ldots$ (2)
$\mathrm{Ut}_{1}+\mathrm{Ut}_{2}=\frac{1}{2} \mathrm{~g} \mathrm{t}_{2}{ }^{2}-\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}{ }^{2} \quad($ from $1 \& 2)$
$\mathrm{U}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \times\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)$
$\mathrm{U}=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)$
$\begin{aligned}
\therefore & \text { Max height }=\mathrm{h}+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=(\mathrm{h})+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=\mathrm{U} \mathrm{t}_{1}+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}^{2}+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=\frac{\mathrm{g}}{2} \mathrm{t}_{1}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}^{2}+\frac{1}{2 \mathrm{~g}} \frac{\mathrm{g}^{2}}{4}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2} \\
&=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \mathrm{t}_{1}+\frac{\mathrm{gt}_{1}{ }^{2}}{2}+\frac{\mathrm{g}}{8}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2} \\
&=\left(\frac{\mathrm{g}}{2}\right)\left\{\left(\mathrm{t}_{2} \mathrm{t}_{1}-\mathrm{t}_{1}{ }^{2}\right)+\mathrm{t}_{1}{ }^{2}+\frac{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2}}{4}\right\} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left\{4 \mathrm{t}_{2} \mathrm{t}_{1}-4 \mathrm{t}_{1}{ }^{2}+4 \mathrm{t}_{1}{ }^{2}+\mathrm{t}_{2}^{2}+\mathrm{t}_{1}^{2}-2 \mathrm{t}_{1} \mathrm{t}_{2}\right\}}{4} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left\{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+2 \mathrm{t}_{1} \mathrm{t}_{2}\right\}}{4} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}}{4} \\
&=\frac{\mathrm{g}}{8}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}
\end{aligned}$
Hence correct Answer is (B).
$\mathrm{~h}=\mathrm{Ut}_{1}+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}{ }^{2} \ldots .$(1)
$-\mathrm{h}=\mathrm{Ut}_{2}-\frac{1}{2} \mathrm{~g} \mathrm{t}_{2}{ }^{2} \ldots \ldots$ (2)
$\mathrm{Ut}_{1}+\mathrm{Ut}_{2}=\frac{1}{2} \mathrm{~g} \mathrm{t}_{2}{ }^{2}-\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}{ }^{2} \quad($ from $1 \& 2)$
$\mathrm{U}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \times\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)$
$\mathrm{U}=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)$
$\begin{aligned}
\therefore & \text { Max height }=\mathrm{h}+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=(\mathrm{h})+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=\mathrm{U} \mathrm{t}_{1}+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}^{2}+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=\frac{\mathrm{g}}{2} \mathrm{t}_{1}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}^{2}+\frac{1}{2 \mathrm{~g}} \frac{\mathrm{g}^{2}}{4}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2} \\
&=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \mathrm{t}_{1}+\frac{\mathrm{gt}_{1}{ }^{2}}{2}+\frac{\mathrm{g}}{8}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2} \\
&=\left(\frac{\mathrm{g}}{2}\right)\left\{\left(\mathrm{t}_{2} \mathrm{t}_{1}-\mathrm{t}_{1}{ }^{2}\right)+\mathrm{t}_{1}{ }^{2}+\frac{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2}}{4}\right\} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left\{4 \mathrm{t}_{2} \mathrm{t}_{1}-4 \mathrm{t}_{1}{ }^{2}+4 \mathrm{t}_{1}{ }^{2}+\mathrm{t}_{2}^{2}+\mathrm{t}_{1}^{2}-2 \mathrm{t}_{1} \mathrm{t}_{2}\right\}}{4} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left\{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+2 \mathrm{t}_{1} \mathrm{t}_{2}\right\}}{4} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}}{4} \\
&=\frac{\mathrm{g}}{8}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}
\end{aligned}$
Hence correct Answer is (B).
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