Question

Are a range of a target is found to be 20 km a share leaves a gun with a velocity of 500 m/s what will be the angle of elevation of the gun if the gun is leveled take g as 10 meters per seconds

Answer 1

To find the angle of elevation of the gun needed to hit a target at a horizontal distance of 20 km with an initial velocity of 500 m/s, we can use the projectile motion equations.

Given:
- Range \( R = 20 \, \text{km} = 20000 \, \text{m} \)
- Initial velocity \( v = 500 \, \text{m/s} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

The formula for the range \( R \) of a projectile launched at an angle \( \theta \) is given by:

\[
R = \frac{v^2 \sin(2\theta)}{g}
\]

Rearranging the formula to find \( \sin(2\theta) \):

\[
\sin(2\theta) = \frac{R \cdot g}{v^2}
\]

Substituting the given values:

\[
\sin(2\theta) = \frac{20000 \, \text{m} \cdot 10 \, \text{m/s}^2}{(500 \, \text{m/s})^2}
\]

Calculating the right side:

\[
\sin(2\theta) = \frac{200000}{250000} = 0.8
\]

Now we need to find \( 2\theta \):

\[
2\theta = \arcsin(0.8)
\]

Calculating \( \arcsin(0.8) \):

\[
2\theta \approx 53.13^\circ
\]

Now, to find \( \theta \):

\[
\theta \approx \frac{53.13^\circ}{2} \approx 26.57^\circ
\]

Therefore, the angle of elevation \( \theta \) at which the gun should be leveled is approximately **26.57 degrees**.