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The problem depicted involves projectile motion and asks for the possible angles and time required for a bullet fired at a given velocity to hit a target located 91.4 meters away. The speed of the bullet is given as \(v_0 = 427 \, \text{m/s}\).

Let's solve part (a) first, where we are asked to find the two possible angles, \(\theta_1\) and \(\theta_2\), that the bullet can be fired at to hit the target. The formula used in projectile motion to find the range \(R\) is:

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

Where:
- \(v_0\) is the initial velocity,
- \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)),
- \(R\) is the horizontal distance or range,
- \(\theta\) is the launch angle.

Given:
- \(R = 91.4 \, \text{m}\),
- \(v_0 = 427 \, \text{m/s}\),
- \(g = 9.8 \, \text{m/s}^2\).

We can rearrange the formula to solve for \(\theta\):

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

Substituting the values:

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

Now, let me calculate \(\sin(2\theta)\) and find the angles \(\theta_1\) and \(\theta_2\).The value of \(\sin(2\theta)\) is approximately \(0.00491\), which leads to two possible angles:

- \(\theta_1 \approx 0.14^\circ\)
- \(\theta_2 \approx 89.86^\circ\)

These two angles represent the possible launch angles that will allow the bullet to hit the target located 91.4 meters away.

Now, for part (b), we need to calculate the time of flight for both angles. The formula for the time of flight \(T\) in projectile motion is:

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

Let's calculate the time for both angles \(\theta_1\) and \(\theta_2\).The time of flight for the bullet to reach the target is:

- For \(\theta_1 \approx 0.14^\circ\), \(T_1 \approx 0.214\) seconds.
- For \(\theta_2 \approx 89.86^\circ\), \(T_2 \approx 87.14\) seconds.

Thus, the first angle corresponds to a short and direct trajectory, while the second angle gives a much longer and indirect path.

Do you want more details or have any other questions?

Here are 5 related questions to deepen your understanding:
1. How does the angle of launch affect the time of flight in projectile motion?
2. Why are there two possible angles that result in the same range for a projectile?
3. What would happen to the possible angles if the target distance increased?
4. How does air resistance influence the real-world trajectory of a projectile?
5. How would the problem change if the gun were aimed upward from a height?

**Tip:** In projectile motion, for any given range, there are always two angles that produce the same horizontal distance, but with different trajectories and times of flight.

Gambar di atas menunjukkan pandangan dari sebuah senapan yang telah 'diarahkan' untuk mengenai target pada jarak 91,4 meter. Jika kecepatan peluru adalah v₀ = 427 m/s. a) Berapakah dua sudut yang mungkin, yaitu sudut θ₁ dan θ₂, antara laras senapan dan garis horizontal sehingga peluru dapat mengenai target? b) Berapakah waktu yang dibutuhkan peluru untuk mencapai target, dengan menggunakan 2 sudut tersebut?

Learn how to solve a projectile motion problem involving a bullet fired at a speed of 427 m/s to hit a target 91.4 meters away. This problem covers key physics concepts such as the range formula, time of flight, and trigonometric identities.

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