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.