Question

question 2 of 25 you drop a rock down a well that is 31 m deep. how long does it take the rock to hit the bottom of the well? a. 2.5 s b. 1.9 s c. 6.3 s d. 2.7 s

Explanation:

Step1: Identify the kinematic - equation

The rock is in free - fall, and we use the equation $h = v_0t+\frac{1}{2}gt^2$. Since the rock is dropped ($v_0 = 0$), the equation simplifies to $h=\frac{1}{2}gt^2$, where $h$ is the height, $g = 9.8\ m/s^2$ is the acceleration due to gravity, and $t$ is the time.

Step2: Solve for time $t$

From $h=\frac{1}{2}gt^2$, we can re - arrange for $t$:
\[

$$\begin{align*} t&=\sqrt{\frac{2h}{g}}\\ \end{align*}$$

\]
Substitute $h = 31\ m$ and $g = 9.8\ m/s^2$ into the formula:
\[

$$\begin{align*} t&=\sqrt{\frac{2\times31}{9.8}}\\ &=\sqrt{\frac{62}{9.8}}\\ &\approx\sqrt{6.327}\\ &\approx 2.5\ s \end{align*}$$

\]

Answer:

A. 2.5 s