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A stone dropped from the top of a tower of height 300 $\mathrm{m}$ high splashes into the water of a pond near the base of the tower. When is the splash heard at the top? [Speed of sound in air $=340 \mathrm{~m} / \mathrm{s}$ and $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$ ]
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Here, $h=300 \mathrm{~m}, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2, v=340 \mathrm{~m} / \mathrm{s}$
Let $t_1=$ time taken by stone to strike the surface of water
$$
\begin{aligned}
&\text { Using } s=u t_1+\frac{1}{2} g t_1^2 \\
&\Rightarrow 300=0+\frac{1}{2} \times 9.8 \times t_1^2 \\
&\therefore \quad t_1=\sqrt{\frac{300}{4.9}}=7.82 \mathrm{~s} .
\end{aligned}
$$
And, $t_2=$ time taken by sound to reach to top of tower $=\frac{h}{v}=\frac{300}{340}=0.88 \mathrm{~s}$.
$\therefore$ Total time after which the splash is heard $=t_1+$ $t_2=7.82+0.88=8.70 \mathrm{~s}$.
Let $t_1=$ time taken by stone to strike the surface of water
$$
\begin{aligned}
&\text { Using } s=u t_1+\frac{1}{2} g t_1^2 \\
&\Rightarrow 300=0+\frac{1}{2} \times 9.8 \times t_1^2 \\
&\therefore \quad t_1=\sqrt{\frac{300}{4.9}}=7.82 \mathrm{~s} .
\end{aligned}
$$
And, $t_2=$ time taken by sound to reach to top of tower $=\frac{h}{v}=\frac{300}{340}=0.88 \mathrm{~s}$.
$\therefore$ Total time after which the splash is heard $=t_1+$ $t_2=7.82+0.88=8.70 \mathrm{~s}$.
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