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For this projectile motion problem, let's solve it step by step.

### Given dataFor this projectile motion problem, let's solve it step by step.

### Given 
- The bullet is fired at an initial speed \( v_0 = 427 \, \text{m/s} \).
- The horizontal distance to the target is \( x = 91.4 \, \text{m} \).
- Acceleration due to gravity is \( g = 9.8 \, \text{m/s}^2 \).

### Part (a): Finding the possible angles \( \theta_1 \) and \( \theta_2 \)

Using the formula for the range of projectile motion:
\[
R = \frac{v_0^2 \sin(2\theta)}{g}
\]
where:
- \( R \) is the range or the horizontal distance,
- \( v_0 \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

Rearrange the equation to solve for \( \theta \):
\[
\sin(2\theta) = \frac{R \cdot g}{v_0^2}
\]
Substituting the given values:
\[
\sin(2\theta) = \frac{91.4 \times 9.8}{427^2}
\]
Let me calculate this value.

We have:
\[
\sin(2\theta) = 0.00491
\]

Now, to find the two possible angles, we solve for \( 2\theta \) using the inverse sine function:
\[
2\theta = \sin^{-1}(0.00491)
\]
Then divide by 2 to get \( \theta \).

Let me calculate the angles for you.The two possible angles are:
- \( \theta_1 \approx 0.14^\circ \)
- \( \theta_2 \approx 89.86^\circ \)

### Part (b): Time of flight for both angles

To find the time of flight, we use the horizontal motion formula:
\[
t = \frac{R}{v_0 \cos(\theta)}
\]
We will calculate the time for both \( \theta_1 \) and \( \theta_2 \). Let me proceed with that calculation.The time of flight for the two angles is:
- For \( \theta_1 \approx 0.14^\circ \), the time is \( t_1 \approx 0.214 \, \text{seconds} \).
- For \( \theta_2 \approx 89.86^\circ \), the time is \( t_2 \approx 87.14 \, \text{seconds} \).

These results show that the low-angle shot (\( \theta_1 \)) results in a short time to reach the target, while the high-angle shot (\( \theta_2 \)) takes a much longer time.

Would you like further details or clarifications? Here are some related questions:

1. What would happen to the angles if the speed of the bullet was doubled?
2. How does air resistance affect the actual time of flight in a real scenario?
3. What other factors could influence the accuracy of the shot over such a long distance?
4. Can the two angles be used in practical applications for shooting targets at different ranges?
5. How does gravity impact projectiles at different elevations?

**Tip:** For optimal shooting angles in real-life scenarios, using angles between 30° and 45° often results in the best balance between distance and speed.

Solve a projectile motion problem where a bullet is fired at an initial speed of 427 m/s to hit a target 91.4 meters away. Learn how to calculate two possible angles and the corresponding time of flight using trigonometry and kinematic equations. Suitable for high school students in Grades 11-12.

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