To determine how far the boulder dropped (denoted as \( \Delta y \)), we need to use the principles of projectile motion. Here’s a step-by-step process to solve this:
1. **Given Data:**
- Initial horizontal velocity, \( v_x = 6.89 \) m/s
- Horizontal distance traveled, \( \Delta x = 23.5 \) m
2. **Horizontal Motion Analysis:**
- Assuming no air resistance, the horizontal velocity remains constant.
- Time of flight, \( t \) can be calculated from the horizontal motion:
\[
t = \frac{\Delta x}{v_x}
\]
2. Insert the given values:
\[
t = \frac{23.5 \, \text{m}}{6.89 \, \text{m/s}}
\]
\[
t = 3.41 \, \text{s}
\]
3. **Vertical Motion Analysis:**
- The boulder is subjected to gravitational acceleration, \( g = 9.81 \) m/s².
- It starts with an initial vertical velocity of 0 m/s.
- The vertical distance fallen ( \( \Delta y \) ) can be calculated using the equation for the displacement under constant acceleration:
\[
\Delta y = \frac{1}{2} g t^2
\]
4. Insert the known values into the equation:
\[
\Delta y = \frac{1}{2} \cdot 9.81 \, \text{m/s}^2 \cdot (3.41 \, \text{s})^2
\]
5. Perform the calculation:
\[
\Delta y = \frac{1}{2} \cdot 9.81 \cdot 11.6281 \, \text{m}^2/\text{s}^2
\]
\[
\Delta y = \frac{1}{2} \cdot 114.009561 \, \text{m}
\]
\[
\Delta y = 57.0047805 \, \text{m}
\]
\[
\Delta y \approx 57.0 \, \text{m}
\]
So, the boulder dropped approximately 57.0 meters.
a boulder moving horizontally at 6.89 m/s rolls off of a cliff. It lands 23.5m from that base of the cliff. How far did the boulder drop?
triangle y=[?]m
1 answer