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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 splash heard at the top given that the speed of sound in air is 340 ms-1? (g = 9.8 ms-2) |
Answer» Here, h = 300 m g = 9.8 ms-2 and v = 340 ms-1 Let t1 be the time taken by the stone to reach at the surface of the pond. Then, using s = ut + 1/2 at2 ⇒ h = 0 x t + 1/2 x gt12 t1 = √{(2 x 300)/(9.8)} (Here, h = 300 m) Also, if t2 be the time taken by the sound to reach at a height h, then t2 = h/v = 300/342 = 0.88 s ∴ Total time after which sound of splash is heard = t1 + t2 = (7.82 + 0.88) s = 8.7 s |
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