Asked by Emmanuella
A range of a target is found to be 20 km a share leaves a gun with a velocity of 500 m/s what must be the angle of evolution of the gun if the ground is level take g as 10 m/s
Answers
Answered by
Emmanuella
Explain further pls I don't understand
Answered by
GPT-4o mini
To find the angle of elevation for a projectile that reaches a target distance (range) of 20 km with an initial velocity of 500 m/s, we'll use the formula for the range of a projectile fired from a level surface. The formula is:
\[
R = \frac{v^2 \sin(2\theta)}{g}
\]
where:
- \( R \) is the range (20 km = 20,000 m),
- \( v \) is the initial velocity (500 m/s),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( \theta \) is the angle of elevation.
Rearranging the formula to solve for \( \sin(2\theta) \):
\[
\sin(2\theta) = \frac{R \cdot g}{v^2}
\]
Substituting the known values into the equation:
\[
\sin(2\theta) = \frac{20000 \cdot 10}{500^2}
\]
Calculating the right-hand side:
\[
\sin(2\theta) = \frac{200000}{250000} = 0.8
\]
Now we need to find \( 2\theta \) such that:
\[
2\theta = \arcsin(0.8)
\]
Using a calculator to find \( \arcsin(0.8) \):
\[
2\theta \approx 53.13^{\circ}
\]
Now, dividing by 2 to solve for \( \theta \):
\[
\theta \approx \frac{53.13^{\circ}}{2} \approx 26.57^{\circ}
\]
Thus, the angle of elevation \( \theta \) is approximately:
\[
\theta \approx 26.57^{\circ}
\]
\[
R = \frac{v^2 \sin(2\theta)}{g}
\]
where:
- \( R \) is the range (20 km = 20,000 m),
- \( v \) is the initial velocity (500 m/s),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( \theta \) is the angle of elevation.
Rearranging the formula to solve for \( \sin(2\theta) \):
\[
\sin(2\theta) = \frac{R \cdot g}{v^2}
\]
Substituting the known values into the equation:
\[
\sin(2\theta) = \frac{20000 \cdot 10}{500^2}
\]
Calculating the right-hand side:
\[
\sin(2\theta) = \frac{200000}{250000} = 0.8
\]
Now we need to find \( 2\theta \) such that:
\[
2\theta = \arcsin(0.8)
\]
Using a calculator to find \( \arcsin(0.8) \):
\[
2\theta \approx 53.13^{\circ}
\]
Now, dividing by 2 to solve for \( \theta \):
\[
\theta \approx \frac{53.13^{\circ}}{2} \approx 26.57^{\circ}
\]
Thus, the angle of elevation \( \theta \) is approximately:
\[
\theta \approx 26.57^{\circ}
\]
Answered by
GPT-4o mini
Sure! Let's break down the problem step by step.
### Problem Restatement:
- You want to find the angle at which a gun should be fired so that the projectile (like a bullet) lands exactly 20 kilometers away (20,000 meters).
- The bullet leaves the gun with an initial velocity of 500 m/s.
- We will use Earth's gravitational acceleration (\(g\)) as 10 m/s².
### Understanding the Formula:
The range \(R\) of a projectile launched at an angle \(\theta\) with an initial velocity \(v\) on level ground can be calculated using the formula:
\[
R = \frac{v^2 \sin(2\theta)}{g}
\]
- \(R\) is the horizontal distance the projectile travels (20,000 m in this case).
- \(v\) is the initial velocity (500 m/s here).
- \(\theta\) is the angle of launch (what we're trying to find).
- \(g\) is the acceleration due to gravity (10 m/s²).
### Step-by-Step Calculation:
1. **Substituting Values:**
We know \(R = 20000\) m, \(v = 500\) m/s, and \(g = 10\) m/s². Plugging these into the range equation:
\[
20000 = \frac{(500)^2 \sin(2\theta)}{10}
\]
2. **Simplifying the Equation:**
First, calculate \(500^2\) which is \(250000\). So we have:
\[
20000 = \frac{250000 \sin(2\theta)}{10}
\]
Now divide \(250000\) by \(10\):
\[
20000 = 25000 \sin(2\theta)
\]
3. **Next, Isolate \(\sin(2\theta)\):**
\[
\sin(2\theta) = \frac{20000}{25000} = 0.8
\]
4. **Using the Inverse Sine Function:**
Now we need to find \(2\theta\):
\[
2\theta = \arcsin(0.8)
\]
You can use a scientific calculator or a trigonometric table to find that:
\[
2\theta \approx 53.13^{\circ}
\]
5. **Solving for \(\theta\):**
Finally, divide by 2 to find \(\theta\):
\[
\theta \approx \frac{53.13^{\circ}}{2} \approx 26.57^{\circ}
\]
### Conclusion:
The angle at which the gun should be elevated to hit a target exactly 20 km away with an initial velocity of 500 m/s is approximately **26.57 degrees**.
If you have any specific part of the explanation you’d like to elaborate on or further clarify, feel free to ask!
### Problem Restatement:
- You want to find the angle at which a gun should be fired so that the projectile (like a bullet) lands exactly 20 kilometers away (20,000 meters).
- The bullet leaves the gun with an initial velocity of 500 m/s.
- We will use Earth's gravitational acceleration (\(g\)) as 10 m/s².
### Understanding the Formula:
The range \(R\) of a projectile launched at an angle \(\theta\) with an initial velocity \(v\) on level ground can be calculated using the formula:
\[
R = \frac{v^2 \sin(2\theta)}{g}
\]
- \(R\) is the horizontal distance the projectile travels (20,000 m in this case).
- \(v\) is the initial velocity (500 m/s here).
- \(\theta\) is the angle of launch (what we're trying to find).
- \(g\) is the acceleration due to gravity (10 m/s²).
### Step-by-Step Calculation:
1. **Substituting Values:**
We know \(R = 20000\) m, \(v = 500\) m/s, and \(g = 10\) m/s². Plugging these into the range equation:
\[
20000 = \frac{(500)^2 \sin(2\theta)}{10}
\]
2. **Simplifying the Equation:**
First, calculate \(500^2\) which is \(250000\). So we have:
\[
20000 = \frac{250000 \sin(2\theta)}{10}
\]
Now divide \(250000\) by \(10\):
\[
20000 = 25000 \sin(2\theta)
\]
3. **Next, Isolate \(\sin(2\theta)\):**
\[
\sin(2\theta) = \frac{20000}{25000} = 0.8
\]
4. **Using the Inverse Sine Function:**
Now we need to find \(2\theta\):
\[
2\theta = \arcsin(0.8)
\]
You can use a scientific calculator or a trigonometric table to find that:
\[
2\theta \approx 53.13^{\circ}
\]
5. **Solving for \(\theta\):**
Finally, divide by 2 to find \(\theta\):
\[
\theta \approx \frac{53.13^{\circ}}{2} \approx 26.57^{\circ}
\]
### Conclusion:
The angle at which the gun should be elevated to hit a target exactly 20 km away with an initial velocity of 500 m/s is approximately **26.57 degrees**.
If you have any specific part of the explanation you’d like to elaborate on or further clarify, feel free to ask!