To find the depth of the elevator shaft, we need to consider two things: the time it takes for the screwdriver to fall to the bottom of the shaft and the time it takes for the sound of the impact to travel back up to the top of the shaft.
Let's denote:
- \( t_f \) as the time for the screwdriver to fall.
- \( t_s \) as the time for the sound to travel back up.
- The total time until the sound is heard is \( t_f + t_s = 6.0 \) seconds.
Step 1: Calculate the time for the screwdriver to fall
Using the equation for the distance fallen under gravity:
\[ d = \frac{1}{2} g t_f^2 \]
Where:
- \( d \) is the depth of the shaft,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)).
Step 2: Calculate the time for the sound to travel
The sound travels at a speed of approximately \( v_s = 343 , \text{m/s} \) (this speed can vary with temperature, but we will use this value).
The time taken for the sound to travel back up is given by:
\[ t_s = \frac{d}{v_s} \]
Step 3: Set up the equations
We have the two main equations:
- The total time equation: \[ t_f + t_s = 6.0 \quad (1) \]
- The equations for distance: \[ d = \frac{1}{2} g t_f^2 \quad (2) \] \[ t_s = \frac{d}{v_s} \quad (3) \]
Substituting equation (3) into (1): \[ t_f + \frac{d}{v_s} = 6.0 \] Now substituting equation (2) into this equation gives: \[ t_f + \frac{\frac{1}{2} g t_f^2}{v_s} = 6.0 \]
Step 4: Substitute values and solve
Substituting \( g = 9.81 , \text{m/s}^2 \) and \( v_s = 343 , \text{m/s} \):
\[ t_f + \frac{\frac{1}{2} \cdot 9.81 \cdot t_f^2}{343} = 6.0 \]
Multiplying by \( 343 \) to eliminate the denominator:
\[ 343 t_f + \frac{1}{2} \cdot 9.81 \cdot t_f^2 = 6.0 \cdot 343 \]
Calculating \( 6.0 \cdot 343 = 2058 \): \[ 343 t_f + 4.905 t_f^2 = 2058 \]
Rearranging gives: \[ 4.905 t_f^2 + 343 t_f - 2058 = 0 \]
Step 5: Use the quadratic formula
Using the quadratic formula: \[ t_f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 4.905 \), \( b = 343 \), and \( c = -2058 \):
Calculate the discriminant: \[ b^2 - 4ac = 343^2 - 4 \cdot 4.905 \cdot (-2058) \] \[ = 117649 + 40351.24 = 158000.24 \]
Now plug it into the formula: \[ t_f = \frac{-343 \pm \sqrt{158000.24}}{2 \cdot 4.905} \]
Calculating \( \sqrt{158000.24} \approx 397.5 \): \[ t_f = \frac{-343 + 397.5}{9.81} \quad \text{(taking the positive root)} \] \[ t_f \approx \frac{54.5}{9.81} \approx 5.56 , \text{s} \]
Step 6: Find \( d \)
Now we can find \( d \) using equation (2): \[ d = \frac{1}{2} g t_f^2 = \frac{1}{2} \cdot 9.81 \cdot (5.56)^2 \] \[ d \approx \frac{1}{2} \cdot 9.81 \cdot 30.9136 \approx 151.70 , \text{m} \]
Thus, the depth of the elevator shaft is approximately 151.7 meters.