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Question
A stone dropped from the top of a tower of height 300 m high 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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Solution 1
Here, h = 300 m, g = 9.8 ms-2 and velocity of sound, v = 340 ms-1 Let t1be the time taken by the stone to reach at the surface of pond.
Then using `s = ut + 1/2 at^2 1/2 at^2 => h = 0 xx t + 1/2 "gt"_1^2`
`:. t_1 = sqrt((2xx300)/9.8) = 7.82 s`
Also if `t_2` is the time taken by the sound to reach at a height h, then
`t_2 = h/v = 300/340 = 0.88 s`
:. Total time after which sound of splash is heard = `t_1 + t_2`
= 7.82 + 0.88 = 8.7 s
Solution 2
Height of the tower, s = 300 m
Initial velocity of the stone, u = 0
Acceleration, a = g = 9.8 m/s2
Speed of sound in air = 340 m/s
The time (`t_1`) taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:
`s= ut_1 + 1/2 "gt"_1^2`
`300 = 0 + 1/2 xx 9.8 xx t_1^2`
`:. t_1 =sqrt((300xx2)/9.8) = 7.82 s`
Time taken by the sound to reach the top of the tower, `t_2 = 300/340 = 0.88 s`
Therefore, the time after which the splash is heard,`t = t_1 + t_2`
= 7.82 + 0.88 = 8.7 s
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