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

Answer 1

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:

\[ \theta \approx 26.57^{\circ} \]

Answer 2

Explain further pls I don't understand

Answer 3

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!