Question

$\huge{ \boxed{ \red{ \frak{Question :-}}}}$
A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s-¹? (g = 9.8 m s-²)

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Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Height of tower (h) = 300m.
• Acceleration due to gravity (g) = 9.8ms⁻².
• Final velocity (v) = 340ms⁻¹.
• Initial velocity (u) = 0ms⁻¹.

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• When is the splash heard?

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

First, use the second equation of motion to find time:

$\red \bigstar \boxed{\sf{\blue{h=ut+\dfrac{1}{2}gt^2 }}}$

Substituting the values,

$\sf \implies 300=0 \times t+\dfrac{1}{2} \times 9.8 \times (t_{AB})^2$

$\sf \implies \dfrac{2 \times 300}{9.8}= (t_{AB})^2$

$\sf \implies t_{AB} =\sqrt{61.22}$

$\implies \sf{\red{t_{AB}=7.82 \ seconds}}$

Now, we found the time taken from A to B. We need to find the time taken from B to A.

We can use this formula:

$\red \bigstar \boxed{\sf{\blue{Time = \dfrac{Distance}{Speed} }}}$

$\implies \sf t_{BA}=\dfrac{300}{340}$

$\implies \sf{\red{t_{BA} =0.88 \ seconds}}$

Total time = $\sf T_{AB}+T_{BA}$

$\sf \to 7.82 + 0.88$

$\bf{\purple{t= 8.7 \ seconds}} \bigstar$

Time taken = 8.7 seconds.

Three equations of motion:

$\boxed{\substack{\displaystyle \sf v=u+gt \\\\ \displaystyle \sf h=ut+\dfrac{1}{2}gt^2 \\\\ \displaystyle \sf 2gs=v^2-u^2}}$