Based on the image provided, this is a projectile motion problem where a bullet is fired towards a target 91.4 meters away with an initial velocity $v_0 = 427 \, \text{m/s}$. To calculate the time it takes for the bullet to reach the target and analyze the trajectory, we need to consider the following:

The horizontal range formula for projectile motion is:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

Where:

• $R$ is the horizontal range (91.4 meters),
• $v_0$ is the initial velocity (427 m/s),
• $\theta$ is the launch angle,
• $g$ is the acceleration due to gravity (approximately 9.8 m/s²).

To find the angle $\theta_2$, we can solve the equation for $\theta$.

Let me compute the values for the angle.The two possible angles that can be used to hit the target 91.4 meters away are approximately:

• $\theta_1 \approx 0.14^\circ$
• $\theta_2 \approx 89.86^\circ$

These two angles represent different paths the projectile can take to reach the same target: one is a very shallow angle, and the other is almost vertical.

Would you like more details or have any questions about this? Here are five related questions to consider:

1. How does the horizontal range change if the initial velocity decreases?
2. What is the time of flight for both angles?
3. How can we calculate the maximum height of the projectile?
4. How does air resistance affect the actual path of the bullet?
5. What is the relationship between launch angle and projectile range?

Tip: In projectile motion, two different angles can result in the same range, but the paths taken will differ significantly.