Calculating Launch Angle for Projectile Motion
The image contains a text in Italian that talks about a projectile being fired with an initial velocity, and it asks to calculate the launch angle such that the projectile can hit a target placed a certain distance away. Unfortunately, the full details of the problem are not completely visible, but I can provide a generic approach for such a problem.
To solve for the launch angle (\(\theta\)) of a projectile given the initial velocity (\(v_0\)) and the distance to the target (\(R\)), we can use the following formula, assuming no air resistance and that the launch and target points are at the same vertical level:
\[ R = \frac{v_0^2}{g} \sin(2 \theta) \]
Where \( R \) is the range or distance to the target, \( v_0 \) is the initial velocity, \( g \) is the acceleration due to gravity (which is approximately \( 9.81 m/s^2 \) on Earth), and \( \theta \) is the launch angle.
Rearrange this equation to solve for \( \theta \):
\[ \sin(2 \theta) = \frac{R g}{v_0^2} \]
Take the inverse sine to find \( 2\theta \):
\[ 2\theta = \sin^{-1}\left(\frac{R g}{v_0^2}\right) \]
Then, divide by 2 to find the angle \( \theta \):
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{R g}{v_0^2}\right) \]
Now, input the given values (\( v_0 = 130 m/s \) and \( R = 1100 m \)), and solve for \( \theta \):
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{1100 \times 9.81}{(130)^2}\right) \]
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{10791}{16900}\right) \]
\[ \theta = \frac{1}{2} \sin^{-1}(0.6383) \]
\[ \theta \approx \frac{1}{2} \times 39.86^\circ \]
\[ \theta \approx 19.93^\circ \]
The launch angle needed to hit the target 1100 meters away with an initial velocity of 130 m/s is approximately \( 19.93^\circ \). Keep in mind that due to the nature of the sin function there are two possible solutions for \( \theta \) in the range [0°, 90°], as sin(θ) = sin(180° - θ). Therefore, the other possible angle would be \( 90^\circ - 19.93^\circ = 70.07^\circ \). These two angles represent the low and high trajectories that both reach the target.
