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This problem involves **projectile motion** where we need to calculate the two possible angles, \( \theta_1 \) and \( \theta_2 \), at which a bullet with an initial velocity \( v_0 = 427 \, \text{m/s} \) can be shot to hit a target 91.4 meters away.

The general equation for the horizontal range \( R \) in projectile motion is given by:

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

Where:
- \( R = 91.4 \, \text{m} \) (the horizontal range to the target),
- \( v_0 = 427 \, \text{m/s} \) (initial velocity),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( \theta \) is the angle at which the bullet is shot.

### Step 1: Solve for \( \sin(2\theta) \)
We can rearrange the equation to solve for \( \sin(2\theta) \):

\[
\sin(2\theta) = \frac{gR}{v_0^2}
\]

Substituting the values:

\[
\sin(2\theta) = \frac{9.8 \times 91.4}{427^2}
\]

I'll compute this value first.We have \( \sin(2\theta) = 0.00491 \).

### Step 2: Find the two possible values for \( 2\theta \)
Since \( \sin(2\theta) \) gives two possible angles for \( 2\theta \) (due to the periodic nature of the sine function), we use:

\[
2\theta_1 = \sin^{-1}(0.00491)
\]
\[
2\theta_2 = 180^\circ - 2\theta_1
\]

I'll calculate these values now.The two possible values for \( 2\theta \) are:
- \( 2\theta_1 \approx 0.28^\circ \)
- \( 2\theta_2 \approx 179.72^\circ \)

### Step 3: Solve for \( \theta_1 \) and \( \theta_2 \)
Now, divide these by 2 to get the two possible angles \( \theta_1 \) and \( \theta_2 \):

\[
\theta_1 = \frac{2\theta_1}{2}, \quad \theta_2 = \frac{2\theta_2}{2}
\]

I'll calculate these angles.The two possible angles are:
- \( \theta_1 \approx 0.14^\circ \)
- \( \theta_2 \approx 89.86^\circ \)

One of these angles (\( 89.86^\circ \)) is so steep that it is impractical for target shooting, which is why it is typically not used.

Would you like more details or further clarification?

### 5 Related Questions:
1. How does air resistance affect the actual trajectory in real-world projectile motion?
2. What are the effects of varying the initial velocity on the range of the projectile?
3. Can you explain how to calculate the maximum height of the projectile?
4. How would you determine the time of flight for the projectile at each angle?
5. How would this calculation change if the target were at a different height?

### Tip:
In projectile motion, always check whether you need to consider both the horizontal and vertical components of motion separately.

The figure shows a rifle that has been 'sighted in' for a 91.4-meter target. If the muzzle speed of the bullet is v₀ = 427 m/s, what are the two possible angles θ₁ and θ₂ between the rifle barrel and the horizontal such that the bullet will hit the target? One of these angles is so large that it is never used in target shooting.

Learn how to calculate the two possible angles θ₁ and θ₂ for a bullet shot from a rifle with a muzzle speed of 427 m/s to hit a target 91.4 meters away. This problem involves key concepts from projectile motion and trigonometry.

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