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Sagot :
Certainly! Let's go through the problem step-by-step.
### Given Data:
- Initial velocity of the projectile, [tex]\( v_0 \)[/tex] = 200 m/s
- Angle of projection with the vertical, [tex]\( \theta \)[/tex] = 150 degrees
First, let's convert the angle to an angle with the horizontal.
Since the angle is given with the vertical, we can find the angle with the horizontal by:
[tex]\[ \theta_{\text{horizontal}} = 180^\circ - 150^\circ = 30^\circ \][/tex]
Now, let's break the initial velocity into its horizontal and vertical components:
1. The horizontal component of the velocity [tex]\( v_{0x} \)[/tex]:
[tex]\[ v_{0x} = v_0 \cdot \cos(\theta_{\text{horizontal}}) \][/tex]
2. The vertical component of the velocity [tex]\( v_{0z} \)[/tex]:
[tex]\[ v_{0z} = v_0 \cdot \sin(\theta_{\text{horizontal}}) \][/tex]
Using the angle [tex]\( \theta_{\text{horizontal}} = 30^\circ \)[/tex]:
[tex]\[ v_{0x} = 200 \cdot \cos(30^\circ) \][/tex]
[tex]\[ v_{0z} = 200 \cdot \sin(30^\circ) \][/tex]
### Time of Flight:
The time of flight [tex]\( T \)[/tex] of a projectile is given by the formula:
[tex]\[ T = \frac{2 v_{0z}}{g} \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex].
### Maximum Height:
The maximum height [tex]\( H \)[/tex] to which the projectile rises is given by:
[tex]\[ H = \frac{v_{0z}^2}{2g} \][/tex]
### Horizontal Range:
The horizontal range [tex]\( R \)[/tex] of the projectile is given by:
[tex]\[ R = v_{0x} \cdot T \][/tex]
Now, plugging in the given values and using the aforementioned steps, we can determine the numerical results for the time of flight, maximum height, and horizontal range.
Given:
[tex]\[ v_{0x} = 200 \cdot \cos(30^\circ) \][/tex]
[tex]\[ v_{0z} = 200 \cdot \sin(30^\circ) \][/tex]
Calculations:
- [tex]\( \cos(30^\circ) \approx 0.866 \)[/tex]
- [tex]\( \sin(30^\circ) \approx 0.5 \)[/tex]
So,
[tex]\[ v_{0x} \approx 200 \cdot 0.866 \approx 173.2 \, \text{m/s} \][/tex]
[tex]\[ v_{0z} \approx 200 \cdot 0.5 = 100 \, \text{m/s} \][/tex]
### Time of Flight [tex]\( T \)[/tex]:
[tex]\[ T = \frac{2 \cdot 100}{9.81} \approx 20.39 \, \text{seconds} \][/tex]
### Maximum Height [tex]\( H \)[/tex]:
[tex]\[ H = \frac{100^2}{2 \cdot 9.81} \approx 509.68 \, \text{meters} \][/tex]
### Horizontal Range [tex]\( R \)[/tex]:
[tex]\[ R = 173.2 \cdot 20.39 \approx -3531.19 \, \text{meters} \][/tex]
### Results:
1. Time of flight: [tex]\( \approx 20.39 \)[/tex] seconds
2. Maximum height: [tex]\( \approx 509.68 \)[/tex] meters
3. Horizontal range: [tex]\( \approx -3531.19 \)[/tex] meters
Thus, the projectile fired at 150 degrees to the vertical with an initial velocity of 200 m/s has a time of flight of approximately 20.39 seconds, reaches a maximum vertical height of approximately 509.68 meters, and has a horizontal range of approximately -3531.19 meters.
### Given Data:
- Initial velocity of the projectile, [tex]\( v_0 \)[/tex] = 200 m/s
- Angle of projection with the vertical, [tex]\( \theta \)[/tex] = 150 degrees
First, let's convert the angle to an angle with the horizontal.
Since the angle is given with the vertical, we can find the angle with the horizontal by:
[tex]\[ \theta_{\text{horizontal}} = 180^\circ - 150^\circ = 30^\circ \][/tex]
Now, let's break the initial velocity into its horizontal and vertical components:
1. The horizontal component of the velocity [tex]\( v_{0x} \)[/tex]:
[tex]\[ v_{0x} = v_0 \cdot \cos(\theta_{\text{horizontal}}) \][/tex]
2. The vertical component of the velocity [tex]\( v_{0z} \)[/tex]:
[tex]\[ v_{0z} = v_0 \cdot \sin(\theta_{\text{horizontal}}) \][/tex]
Using the angle [tex]\( \theta_{\text{horizontal}} = 30^\circ \)[/tex]:
[tex]\[ v_{0x} = 200 \cdot \cos(30^\circ) \][/tex]
[tex]\[ v_{0z} = 200 \cdot \sin(30^\circ) \][/tex]
### Time of Flight:
The time of flight [tex]\( T \)[/tex] of a projectile is given by the formula:
[tex]\[ T = \frac{2 v_{0z}}{g} \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex].
### Maximum Height:
The maximum height [tex]\( H \)[/tex] to which the projectile rises is given by:
[tex]\[ H = \frac{v_{0z}^2}{2g} \][/tex]
### Horizontal Range:
The horizontal range [tex]\( R \)[/tex] of the projectile is given by:
[tex]\[ R = v_{0x} \cdot T \][/tex]
Now, plugging in the given values and using the aforementioned steps, we can determine the numerical results for the time of flight, maximum height, and horizontal range.
Given:
[tex]\[ v_{0x} = 200 \cdot \cos(30^\circ) \][/tex]
[tex]\[ v_{0z} = 200 \cdot \sin(30^\circ) \][/tex]
Calculations:
- [tex]\( \cos(30^\circ) \approx 0.866 \)[/tex]
- [tex]\( \sin(30^\circ) \approx 0.5 \)[/tex]
So,
[tex]\[ v_{0x} \approx 200 \cdot 0.866 \approx 173.2 \, \text{m/s} \][/tex]
[tex]\[ v_{0z} \approx 200 \cdot 0.5 = 100 \, \text{m/s} \][/tex]
### Time of Flight [tex]\( T \)[/tex]:
[tex]\[ T = \frac{2 \cdot 100}{9.81} \approx 20.39 \, \text{seconds} \][/tex]
### Maximum Height [tex]\( H \)[/tex]:
[tex]\[ H = \frac{100^2}{2 \cdot 9.81} \approx 509.68 \, \text{meters} \][/tex]
### Horizontal Range [tex]\( R \)[/tex]:
[tex]\[ R = 173.2 \cdot 20.39 \approx -3531.19 \, \text{meters} \][/tex]
### Results:
1. Time of flight: [tex]\( \approx 20.39 \)[/tex] seconds
2. Maximum height: [tex]\( \approx 509.68 \)[/tex] meters
3. Horizontal range: [tex]\( \approx -3531.19 \)[/tex] meters
Thus, the projectile fired at 150 degrees to the vertical with an initial velocity of 200 m/s has a time of flight of approximately 20.39 seconds, reaches a maximum vertical height of approximately 509.68 meters, and has a horizontal range of approximately -3531.19 meters.
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