A janitor standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from rest and hear it hit bottom after 2.42 s. (a) How far (in m) is it from where they drop the rock to the bottom of the shaft? The speed of sound at the temperature of air in the shaft is 336 m/s. (Round your answer to at least three significant figures. Use g = 9.80 m/s2 as needed.) m (b) If the travel time for the sound is ignored, what percent error is introduced in the determination of depth of the shaft? %

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aristocles aristocles · Answer 1 · 2019-09-15T06:53:00+02:00

Answer:

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

[tex]t_1 = \sqrt{\frac{2H}{g}}[/tex]

now the sound will come back to the observer in the time

[tex]t_2 = \frac{H}{v}[/tex]

so we will have

[tex]t_1 + t_2 = 2.42[/tex]

[tex]\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42[/tex]

so we have

[tex]\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42[/tex]

solve above equation for H

[tex]H = 26.8 m[/tex]

Part b)

If sound reflection part is ignored then in that case

[tex]H = \frac{1}{2}gt^2[/tex]

[tex]H = \frac{1}{2}(9.81)(2.42^2)[/tex]

[tex]H = 28.7 m[/tex]

so here percentage error in height calculation is given as

[tex]percentage = \frac{28.7 - 26.8}{26.8} \times 100[/tex]

[tex]percentage = 7.18 [/tex]