At Zion National Park a loud shout produces an echo 10.50 s later from a colorfu

Question

At Zion National Park a loud shout produces an echo 10.50 s later from a colorful sandstone cliff. How far away is the cliff? is within 10% of the correct value. This may be due to round off error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize round off error, m When you drop a rock into a well you hear the splash 2.0 seconds later, How deep is the well? If the depth of the well were doubled, would the time required to hear the splash be greater than, less than, or equal to 4.0 seconds?

Explanation / Answer

echo is produced after 10.5 s

speed of sound is 343 m/s

distance = speed*time

2d = vt

2d = 343*10.5

d = 1800.75 m

the cliff is 1800.75 m far away

b) time taken for the rock to reach the water

h = gt^2/2

t = sqrt(2h/g)

then sound of splash take a time h/v to travel back

total time = sqrt(2h/g) + h/v

let sqrt h = x

2 = x*sqrt(2/g) + x^2/v

x^2/343 + 0.45x - 2 = 0

x = -0.45 + sqrt(0.45^2 + 4*2/343) / 2/343

x = 4.436 m

sqrt h = 4.436

h = 19.684

the well is 19.684 m

b) time taken when depth is doubled

h in this case = 2*19.684 = 39.37 m

t =sqrt(2*39.37 / 9.8) + 39.37/343

t = 2.95 s

the time taken is less than 4 seconds

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