A bucket is thrown vertically down a well with a velocity of 4.19 m/s.  The bucket then takes 1.07 s to reach the bottom of the well.  What is the depth of the well?

 

Only give the numerical answer. No units.

Use at least 4 decimals in the procedure.

Provide only 2 decimals in the final result, without rounding.

To find the depth of the well, we use the equation for displacement under constant acceleration:

\[
d = v_0 t + \frac{1}{2} a t^2
\]

Where:
- \( v_0 = 4.19 \, \text{m/s} \) is the initial velocity,
- \( t = 1.07 \, \text{seconds} \) is the time,
- \( a = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.

Now, calculating step by step.

1. \( v_0 t = 4.19 \times 1.07 = 4.4833 \)
2. \( \frac{1}{2} a t^2 = 0.5 \times 9.81 \times (1.07)^2 = 5.61227585 \)

Now, summing them up:

\[
d = 4.4833 + 5.61227585 = 10.09557585
\]

Final result: 10.09



A bucket is thrown vertically down a well with a velocity of 4.19 m/s. The bucket then takes 1.07 s to reach the bottom of the well. What is the depth of the well?

Learn how to calculate the depth of a well using kinematic equations. This problem involves initial velocity, time, and gravitational acceleration. Step-by-step explanation included. Suitable for students in Grades 9-12.

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