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Q15.2:
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 $s^{-2}$)

Solution :

Step 1: Given Variables & Theoretical Foundation

To determine the time interval between the release of the stone and the hearing of the splash at the top, we must decompose the problem into two distinct physical phases: the free-fall of the stone and the upward propagation of the sound wave.

  • Height of the tower ($h$): $300\text{ m}$
  • Acceleration due to gravity ($g$): $9.8\text{ m/s}^2$
  • Speed of sound ($v_s$): $340\text{ m/s}$
  • Initial velocity of the stone ($u$): $0\text{ m/s}$

Step 2: Calculating Time of Descent ($t_1$)

The time taken for the stone to fall is calculated using the kinematic equation of motion for constant acceleration [Per the Second Equation of Motion]:

$h = ut_1 + \frac{1}{2}gt_1^2$

Since $u = 0$, the equation simplifies to:

$300 = 0 + \frac{1}{2}(9.8)t_1^2$

$300 = 4.9t_1^2$

$t_1^2 = \frac{300}{4.9} \approx 61.224$

$t_1 = \sqrt{61.224} \approx 7.825\text{ s}$

Step 3: Calculating Time for Sound to Travel Upward ($t_2$)

Once the stone hits the water, the sound travels back to the top at a constant velocity. We use the definition of constant velocity [Per the Definition of Uniform Motion]:

$t_2 = \frac{\text{distance}}{\text{speed}} = \frac{h}{v_s}$

$t_2 = \frac{300}{340} \approx 0.882\text{ s}$

Step 4: Total Elapsed Time ($T$)

The total time $T$ from the moment the stone is dropped until the sound is heard is the sum of the two time intervals:

$T = t_1 + t_2$

$T = 7.825\text{ s} + 0.882\text{ s} = 8.707\text{ s}$

Visual Representation of Motion

Sound (v=340 m/s) Stone (g=9.8 m/s²) Top (h=300m) Pond/Base

Final Solution: The splash is heard at the top after approximately 8.71 seconds.


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