Q. A stone is dropped into a lake from a tower $500\ metre$ high. The sound of the splash will be heard by the man approximately after
A) 11.5 seconds
B) 21 seconds
C) 10 seconds
D) 14 seconds
Explanation
(a) The time taken by the stone to reach the lake
$t_{1} = \sqrt{\left( \frac{2h}{g} \right)} = \sqrt{\left( \frac{2 \times 500}{10} \right)} = 10\sec{\ (Using\ h = ut + \frac{1}{2}gt^{2})}$
Now time taken by sound from lake to the man
$t_{2} = \frac{h}{v} = \frac{500}{340} \approx 1.5\ \sec$
$\Rightarrow$ Total time $= t_{1} + t_{2} = 10 + 1.5 = 11.5\ \sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.