a stone is dropped into a well of 44m deep and the sound of splash is heard after 3.12 sec.find the speed of the sound in air
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Time taken by the stone to reach the water surface can be calculated using the equation of motion
s = ut + ½ gt2
Here u = 0, s = 44 m , g = 9.8 m/s2
So
t = (2x44/9.8)½
= 2.996 s ≈ 3 s
Therefore time taken by sound to travel this distance = 3.12 - 3 = 0.12 s
Speed of sound in air = distance covered/Time
= 44 m/ (0.12 s)
= 366.7 m/s
s = ut + ½ gt2
Here u = 0, s = 44 m , g = 9.8 m/s2
So
t = (2x44/9.8)½
= 2.996 s ≈ 3 s
Therefore time taken by sound to travel this distance = 3.12 - 3 = 0.12 s
Speed of sound in air = distance covered/Time
= 44 m/ (0.12 s)
= 366.7 m/s
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Answer:
First the stone will go 44 meters down and hit the water surface. Then the splash sound will travel in from the water surface to the point from where stone is dropped. The total time taken is 3.12 second.
Let the stone reached the water surface in the well in t second.
s = ut + 0.5gt²
⇒ 44 = 0 + 0.5×9.81×t²
⇒ 44 = 4.905t²
⇒ t² = 44/4.905 = 8.97
⇒ t = √8.97 = 2.99s
Time taken by sound to travel 44m = 3.12 - 2.99 = 0.13s
speed of sound in air = 44/0.13 = 338.46 m/s
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