Question
How would i find the vertical displacemnt of a projectile fired parrallel to the ground at 7m/s from a height at 3m
Answers
GPT-4o mini
To find the vertical displacement of a projectile fired parallel to the ground from a height, we can analyze the motion due to gravity. Since the projectile is fired horizontally at a speed of \(7 \, \text{m/s}\) from a height of \(3 \, \text{m}\), the only vertical motion it experiences is due to the gravitational force acting on it.
**Vertical motion analysis:**
1. **Initial conditions:**
- Initial vertical position (\(y_0\)) = \(3 \, \text{m}\)
- Initial vertical velocity (\(v_{y0}\)) = \(0 \, \text{m/s}\) (since the projectile is fired horizontally)
- Acceleration due to gravity (\(g\)) = \(9.81 \, \text{m/s}^2\) (downward)
2. **Equation of motion for vertical displacement:**
The vertical displacement (\(y\)) can be calculated using the following kinematic equation:
\[
y = y_0 + v_{y0} t + \frac{1}{2} a t^2
\]
Here, \(a = -g\) (the negative sign indicates downward acceleration).
Therefore:
\[
y = 3 \, \text{m} + 0 + \frac{1}{2} (-9.81 \, \text{m/s}^2) t^2
\]
This simplifies to:
\[
y = 3 \, \text{m} - 4.905 t^2
\]
3. **Finding the time of flight**:
To find the time of flight (\(t\)), we need to determine how long it takes for the projectile to fall vertically from \(3 \, \text{m}\) to \(y = 0 \, \text{m}\):
\[
0 = 3 - 4.905 t^2
\]
Rearranging gives:
\[
4.905 t^2 = 3
\]
\[
t^2 = \frac{3}{4.905} \approx 0.610
\]
\[
t \approx \sqrt{0.610} \approx 0.781 \, \text{s}
\]
4. **Calculating the vertical displacement**:
Since we want the displacement downwards as the projectile reaches the ground \(y = 0\), the vertical displacement (\(\Delta y\)) is:
\[
\Delta y = y_0 - y = 3 \, \text{m} - 0 = 3 \, \text{m}
\]
Thus, the vertical displacement of the projectile as it hits the ground is **3 meters**.
**Vertical motion analysis:**
1. **Initial conditions:**
- Initial vertical position (\(y_0\)) = \(3 \, \text{m}\)
- Initial vertical velocity (\(v_{y0}\)) = \(0 \, \text{m/s}\) (since the projectile is fired horizontally)
- Acceleration due to gravity (\(g\)) = \(9.81 \, \text{m/s}^2\) (downward)
2. **Equation of motion for vertical displacement:**
The vertical displacement (\(y\)) can be calculated using the following kinematic equation:
\[
y = y_0 + v_{y0} t + \frac{1}{2} a t^2
\]
Here, \(a = -g\) (the negative sign indicates downward acceleration).
Therefore:
\[
y = 3 \, \text{m} + 0 + \frac{1}{2} (-9.81 \, \text{m/s}^2) t^2
\]
This simplifies to:
\[
y = 3 \, \text{m} - 4.905 t^2
\]
3. **Finding the time of flight**:
To find the time of flight (\(t\)), we need to determine how long it takes for the projectile to fall vertically from \(3 \, \text{m}\) to \(y = 0 \, \text{m}\):
\[
0 = 3 - 4.905 t^2
\]
Rearranging gives:
\[
4.905 t^2 = 3
\]
\[
t^2 = \frac{3}{4.905} \approx 0.610
\]
\[
t \approx \sqrt{0.610} \approx 0.781 \, \text{s}
\]
4. **Calculating the vertical displacement**:
Since we want the displacement downwards as the projectile reaches the ground \(y = 0\), the vertical displacement (\(\Delta y\)) is:
\[
\Delta y = y_0 - y = 3 \, \text{m} - 0 = 3 \, \text{m}
\]
Thus, the vertical displacement of the projectile as it hits the ground is **3 meters**.