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Revisiting Waves

The sound fro...

Question

The sound from a very high burst of fireworks takes 5 s to arrive at the observer. The burst occurs 1662 m above the observer and travels vertically through two stratifier layers of air, the top one of thickness
$d_{1}$ at $0^{\circ} C$ and the bottom one of thickness $d_{2}$ at $20^{\circ} C$ . Then (assume velocity of sound at $0^{\circ} C$ is 330 m/s)

d_{1} = 342 m

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d_{2} = 1320 m

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d_{1} = 1485 m

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d_{2} = 342 m

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Solution

The correct option is D $d_{2} = 342 m$
Time taken is given by
$T = t_{1} + t_{2} = \frac{d_{1}}{v_{1}} + \frac{d_{2}}{v_{2}} V_{1} = v_{0}^{\circ} c = 330 m / s v_{2} = (330 + .06 t) = 342 m / s d = 1662 m ∴ T = \frac{d_{1}}{330} + \frac{(d - d_{1})}{342} = 5 s \frac{d_{1} (342 - 330)}{330 \times 342} + \frac{d}{342} = 5 s 12 d_{1} = 5 (342 \times 330) - 330 \times 1662 d_{1} = 1320 m d_{2} = 342 m$

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Similar questions

d_{1}

0^{\circ} C

d_{2}

20^{\circ} C

0^{\circ} C

330 m

1.1 {m/s}^{2}

20 s

330 m/s)

^{\circ} C

0^{\circ} C

0^{\circ} C = 330 {m s}^{- 1}

61^{\circ} C

0^{\circ} C = 330 {m s}^{- 1}

1.1 m s^{- 2}

330 m s^{- 1}

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