Question

# A stone is dropped from the top of a tower $500\mathrm{m}$ high into a pond of water at the base of the tower. When is the splash heard at the top? Given, $g=10m{s}^{-2}$ and speed of sound is $340m{s}^{-1}$

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Solution

## Step 1: GivenTower height = $500\mathrm{m}$Speed of sound =$340m{s}^{-1}$Accleration due to gravity = $g=10m{s}^{-2}$Step 2. Find time taken to the stone to reach the tower's foundation.According to the second equation of motion:$\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^{2}$where,$u$ is the initial velocity $t$ is the time taken by an object to reach the final position.$g$is the acceleration due to gravity.Here, $u$ is the stone's initial velocity $u=0\frac{\mathrm{m}}{\mathrm{s}}$ before falling$g=10m/{s}^{2}$ is the acceleration due to gravity.and $S$ is the distance traveled by the stone i.e. $500\mathrm{m}$ which is the tower's heightAcceleration due to gravity, (g) is $10\mathrm{m}/{\mathrm{s}}^{2}$$t={t}_{1}$ is the length of time it took the stone to reach the tower's foundation.Put the above-given values Find the time taken$t={t}_{1}$ by the stone to reach the base:$\mathrm{S}={\mathrm{ut}}_{1}+\frac{1}{2}{{\mathrm{gt}}_{1}}^{2}\phantom{\rule{0ex}{0ex}}⇒500=0×\mathrm{t}1+\frac{1}{2}×10×{\mathrm{t}}_{1}{}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{t}}_{1}^{2}=100\phantom{\rule{0ex}{0ex}}⇒{t}_{1}=10\mathrm{s}$Step 3. Find time taken by the sound of stone slashed in the pond to reach back to the top of the tower.Time taken by the sound of stone slashed in the pond to reach back to the top of the tower (say ${t}_{2}$) is:${t}_{2}=\frac{dis\mathrm{tan}ce}{speedofsound}\phantom{\rule{0ex}{0ex}}=\frac{500}{340}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=1.47s$Step 3. Find total time time taken after the stone is dropped to the sound of splash reached back.Total time take taken after the stone is dropped to the sound of splash reached back: $\begin{array}{rcl}t& =& {t}_{1}+{t}_{2}\\ & ⇒& t=10+1.47\\ & ⇒& t=11.47s\end{array}$Therefore, the splash is heard at the top after $11.47s$.

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