Question

a rock moving horizontally at 2.87 m/s falls off of a cliff. It lands 5.32m from the base of the cliff. How far did the rock drop?
triangle y=[?]m

Answer 1

To determine how far the rock dropped (denoted as \(\Delta y\)), we'll need to use the principles of projectile motion. The key here is to separate the motion into horizontal and vertical components and analyze them independently.

Given:
- The horizontal speed of the rock (\(v_{x}\)) is 2.87 m/s.
- The horizontal distance the rock travels (\(d_{x}\)) is 5.32 m.

First, let's find the time (\(t\)) it takes for the rock to travel the horizontal distance.
The horizontal motion is described by the equation:
\[
d_{x} = v_{x} \cdot t
\]

Rearranging for \(t\):
\[
t = \frac{d_{x}}{v_{x}}
\]

Plug in the given values:
\[
t = \frac{5.32\ \text{m}}{2.87\ \text{m/s}}
\]
\[
t \approx 1.85\ \text{s}
\]

Now, we'll use this time to find the vertical distance (\(\Delta y\)) the rock dropped. The vertical motion is described by the equation for an object in free fall:
\[
\Delta y = \frac{1}{2} g t^2
\]
where \( g \) is the acceleration due to gravity (\(9.81\ \text{m/s}^2 \)).

Substitute \( t \) and \( g \) into the equation:
\[
\Delta y = \frac{1}{2} \cdot 9.81\ \text{m/s}^2 \cdot (1.85\ \text{s})^2
\]
\[
\Delta y = \frac{1}{2} \cdot 9.81\ \text{m/s}^2 \cdot 3.4225\ \text{s}^2
\]
\[
\Delta y = 4.905\ \text{m/s}^2 \cdot 3.4225\ \text{s}^2
\]
\[
\Delta y \approx 16.78\ \text{m}
\]

So, the rock dropped approximately 16.78 meters.