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A screwdriver is dropped from the top of an elevator shaft. Exactly 6.0 seconds later, the sound of the screwdriver hitting bottom is heard. How deep is the shaft? Hint: The distance that a dropped object falls in t seconds is represented by the formula s = 16t2. The speed of sound is 1100 ft/sec. The shaft is approximately (Round to the nearest tenth.) | feet deep.

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Answer:

Step-by-step explanation:

Using the equation of motion

S = ut+1/2gt²

S is the height of fall

g is the acceleration due to gravity = 9.81m/s²

u is the initial velocity = 0m/s

t is the time = 6s

Substitute in the formula;

S = 0+1/2(9.81)×6²

S = 4.905×36

S = 176.58m

The shaft is 176.58m deep

Given the expression for distance

S = 16t²

Velocity V = dS/dt

V = 32t

Given v = 1100

1100 = 32t

t = 1100/32

t = 34.375secs

Substitute t = 34.375 into S = 16t²

S = 16(34.375)²

S = 18,906.25ft

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