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!