Question

select the correct answer.
a stone is dropped from a tower 100 meters above the ground. the stone falls past ground level and into a well. it hits the water at the bottom of the well 5.00 seconds after being dropped from the tower. calculate the depth of the well. given: $g = -9.81$ meters/second$^2$.
a. 22.62 meters
b. 50.7 meters
c. 10 meters
d. 11.25 meters
e. 15 meters

Explanation:

Step1: Recall free fall displacement formula

The displacement of a freely falling object is given by:
$$y = y_0 + v_0 t + \frac{1}{2} g t^2$$
Where:

• $y_0 = 100\ \text{m}$ (initial height above ground),
• $v_0 = 0\ \text{m/s}$ (dropped from rest),
• $g = -9.81\ \text{m/s}^2$,
• $t = 5.00\ \text{s}$,
• $y = -d$ (final position, $d$ is well depth, negative as below ground).

Step2: Substitute known values

$$-d = 100 + 0 + \frac{1}{2}(-9.81)(5.00)^2$$

Step3: Calculate the right-hand side

First compute $\frac{1}{2}(-9.81)(25) = -122.625$
$$-d = 100 - 122.625$$
$$-d = -22.625$$

Step4: Solve for well depth $d$

Multiply both sides by $-1$:
$$d = 22.625 \approx 22.62\ \text{meters}$$

Answer:

A. 22.62 meters