Question
A catapult launches a boulder at 72.0 m/s from the top of a 15.0 m-high castle wall at an angle of 30.0 degrees. What is the boulder's impact speed when it hits the ground below?
Answers
GPT-4o mini
To determine the boulder's impact speed when it hits the ground, we need to break down the problem into vertical and horizontal components and analyze the motion using kinematic equations.
### Step 1: Initial Velocity Components
The initial velocity (\(v_0\)) of the boulder is 72.0 m/s at an angle of 30.0 degrees. We can resolve this velocity into its horizontal and vertical components:
- Horizontal component (\(v_{0x}\)):
\[
v_{0x} = v_0 \cdot \cos(\theta) = 72.0 \cdot \cos(30^\circ) = 72.0 \cdot \frac{\sqrt{3}}{2} \approx 62.35 \, \text{m/s}
\]
- Vertical component (\(v_{0y}\)):
\[
v_{0y} = v_0 \cdot \sin(\theta) = 72.0 \cdot \sin(30^\circ) = 72.0 \cdot \frac{1}{2} = 36.0 \, \text{m/s}
\]
### Step 2: Time of Flight Calculation
To find the time it takes for the boulder to hit the ground, we need to use the vertical motion equation. The boulder is launched from a height of \(h = 15.0 \, \text{m}\) above the ground. The vertical motion can be described by the equation:
\[
y = v_{0y} \cdot t - \frac{1}{2} g t^2
\]
Where:
- \(y\) is the vertical displacement (which will be -15.0 m, since it falls down).
- \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)).
- \(t\) is the time in seconds.
Setting up the equation, we have:
\[
-15.0 = 36.0 \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2
\]
Rearranging gives us:
\[
4.905 t^2 - 36.0 t - 15.0 = 0
\]
### Step 3: Solving the Quadratic Equation
We can use the quadratic formula (\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)) where \(a = 4.905\), \(b = -36.0\), and \(c = -15.0\):
\[
b^2 - 4ac = (-36.0)^2 - 4 \cdot 4.905 \cdot (-15.0) = 1296 + 294.3 = 1590.3
\]
Calculating \(t\):
\[
t = \frac{-(-36.0) \pm \sqrt{1590.3}}{2 \cdot 4.905}
\]
\[
t = \frac{36.0 \pm 39.87}{9.81}
\]
Calculating the two possible values for \(t\):
1. \(t = \frac{36.0 + 39.87}{9.81} \approx \frac{75.87}{9.81} \approx 7.73 \, \text{s}\)
2. \(t = \frac{36.0 - 39.87}{9.81} \approx \frac{-3.87}{9.81}\) (not valid, as time cannot be negative)
Thus, the valid time of flight is approximately \(t \approx 7.73 \, \text{s}\).
### Step 4: Final Vertical Velocity
To find the final vertical velocity (\(v_{fy}\)) when the boulder hits the ground, we can use:
\[
v_{fy} = v_{0y} - g t
\]
Substituting the known values:
\[
v_{fy} = 36.0 - (9.81 \cdot 7.73)
\]
\[
v_{fy} = 36.0 - 75.93 \approx -39.93 \, \text{m/s}
\]
### Step 5: Final Result
The magnitude of the boulder's velocity just before impact can be calculated using the horizontal and vertical components:
Now, we find the final speed (\(v_f\)) using:
\[
v_f = \sqrt{v_{fx}^2 + v_{fy}^2}
\]
Where \(v_{fx} = v_{0x} = 62.35 \, \text{m/s}\) (the horizontal component does not change throughout the flight).
Calculating \(v_f\):
\[
v_f = \sqrt{(62.35)^2 + (-39.93)^2}
\]
\[
v_f = \sqrt{3886.52 + 1595.61} \approx \sqrt{5482.13} \approx 74.1 \, \text{m/s}
\]
### Conclusion
The boulder's impact speed when it hits the ground below is approximately **74.1 m/s**.
### Step 1: Initial Velocity Components
The initial velocity (\(v_0\)) of the boulder is 72.0 m/s at an angle of 30.0 degrees. We can resolve this velocity into its horizontal and vertical components:
- Horizontal component (\(v_{0x}\)):
\[
v_{0x} = v_0 \cdot \cos(\theta) = 72.0 \cdot \cos(30^\circ) = 72.0 \cdot \frac{\sqrt{3}}{2} \approx 62.35 \, \text{m/s}
\]
- Vertical component (\(v_{0y}\)):
\[
v_{0y} = v_0 \cdot \sin(\theta) = 72.0 \cdot \sin(30^\circ) = 72.0 \cdot \frac{1}{2} = 36.0 \, \text{m/s}
\]
### Step 2: Time of Flight Calculation
To find the time it takes for the boulder to hit the ground, we need to use the vertical motion equation. The boulder is launched from a height of \(h = 15.0 \, \text{m}\) above the ground. The vertical motion can be described by the equation:
\[
y = v_{0y} \cdot t - \frac{1}{2} g t^2
\]
Where:
- \(y\) is the vertical displacement (which will be -15.0 m, since it falls down).
- \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)).
- \(t\) is the time in seconds.
Setting up the equation, we have:
\[
-15.0 = 36.0 \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2
\]
Rearranging gives us:
\[
4.905 t^2 - 36.0 t - 15.0 = 0
\]
### Step 3: Solving the Quadratic Equation
We can use the quadratic formula (\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)) where \(a = 4.905\), \(b = -36.0\), and \(c = -15.0\):
\[
b^2 - 4ac = (-36.0)^2 - 4 \cdot 4.905 \cdot (-15.0) = 1296 + 294.3 = 1590.3
\]
Calculating \(t\):
\[
t = \frac{-(-36.0) \pm \sqrt{1590.3}}{2 \cdot 4.905}
\]
\[
t = \frac{36.0 \pm 39.87}{9.81}
\]
Calculating the two possible values for \(t\):
1. \(t = \frac{36.0 + 39.87}{9.81} \approx \frac{75.87}{9.81} \approx 7.73 \, \text{s}\)
2. \(t = \frac{36.0 - 39.87}{9.81} \approx \frac{-3.87}{9.81}\) (not valid, as time cannot be negative)
Thus, the valid time of flight is approximately \(t \approx 7.73 \, \text{s}\).
### Step 4: Final Vertical Velocity
To find the final vertical velocity (\(v_{fy}\)) when the boulder hits the ground, we can use:
\[
v_{fy} = v_{0y} - g t
\]
Substituting the known values:
\[
v_{fy} = 36.0 - (9.81 \cdot 7.73)
\]
\[
v_{fy} = 36.0 - 75.93 \approx -39.93 \, \text{m/s}
\]
### Step 5: Final Result
The magnitude of the boulder's velocity just before impact can be calculated using the horizontal and vertical components:
Now, we find the final speed (\(v_f\)) using:
\[
v_f = \sqrt{v_{fx}^2 + v_{fy}^2}
\]
Where \(v_{fx} = v_{0x} = 62.35 \, \text{m/s}\) (the horizontal component does not change throughout the flight).
Calculating \(v_f\):
\[
v_f = \sqrt{(62.35)^2 + (-39.93)^2}
\]
\[
v_f = \sqrt{3886.52 + 1595.61} \approx \sqrt{5482.13} \approx 74.1 \, \text{m/s}
\]
### Conclusion
The boulder's impact speed when it hits the ground below is approximately **74.1 m/s**.