What's the air temperature when Δd=250m

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To calculate the air temperature based on the echo from a cliff, the distance of 250 m is divided by 2 to find the one-way distance of 125 m. The speed of sound is determined using the time of 1.5 seconds, resulting in a calculated speed of 83.33 m/s. This speed is then set equal to the equation for the speed of sound, v = 331 m/s + (0.59 m/s/°C)T, to solve for temperature T. The discussion raises questions about the distance traveled by the sound wave and the reasoning behind using half the distance. Understanding these calculations is essential for determining the air temperature accurately.
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Homework Statement



A hunter wanted to know the air temperature. The echo from a nearby cliff returned 1.5 s after he fired his riffle. If the cliff is 250 m away, how do you calculate the air temperature?

Homework Equations



v=331 m/s + (0.59 m/s/c˚)T niSinθi=nRSinθR x=Δd/2 These eq'ns might be useful

The Attempt at a Solution



x=250 m/2 = 125
v=125 m/1.5s = 83.33 m/s
83.33 m/s = 331 m/s + (0.59 m/s/c˚)T
 
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why you used x = 125??

what is the distance traveled by sound wave in the whole process?
 

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