What two angles of elevation will enable a projectile to reach a target 13 km downrange on the same level as the gun if the​ projectile's initial speed is 403 ​m/sec?
the two angles of elevation are _° and _°

To solve this problem, we can use the range formula for projectile motion:

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

Where:
- \( R \) is the range (13,000 m),
- \( v \) is the initial velocity (403 m/s),
- \( g \) is the acceleration due to gravity (9.81 m/s²),
- \( \theta \) is the angle of elevation.

We want to find two angles \( \theta_1 \) and \( \theta_2 \) that satisfy this equation.

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

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

Substitute the known values:

\[
\sin(2\theta) = \frac{13,000 \times 9.81}{403^2}
\]

Now, I'll calculate the value of \( \sin(2\theta) \).We have:

\[
\sin(2\theta) = 0.7852
\]

### Step 2: Solve for \( 2\theta \)
Now, take the inverse sine (arcsin) to find \( 2\theta \):

\[
2\theta = \arcsin(0.7852)
\]

I'll compute the value of \( 2\theta \).We have:

\[
2\theta = 51.74^\circ
\]

### Step 3: Solve for \( \theta \)
Now, divide by 2 to find one of the possible angles:

\[
\theta_1 = \frac{51.74^\circ}{2} = 25.87^\circ
\]

However, since \( \sin(2\theta) \) is periodic, there is a second solution. The other angle will be:

\[
2\theta_2 = 180^\circ - 51.74^\circ = 128.26^\circ
\]

Thus:

\[
\theta_2 = \frac{128.26^\circ}{2} = 64.13^\circ
\]

### Final Answer:
The two angles of elevation are approximately \( 25.87^\circ \) and \( 64.13^\circ \).

Would you like any further explanation or details?

---
1. How does the range formula work in projectile motion?
2. What assumptions are we making in these calculations?
3. Can you explain why there are two angles of elevation?
4. What role does the initial speed play in determining the angles?
5. How does gravity affect projectile motion?

**Tip:** The angle for maximum range is always \(45^\circ\) when air resistance is ignored.

What two angles of elevation will enable a projectile to reach a target 13 km downrange on the same level as the gun if the​ projectile's initial speed is 403 ​m/sec? The two angles of elevation are _° and _°

Learn how to calculate the two angles of elevation required for a projectile to travel 13 km downrange at an initial speed of 403 m/sec. This problem involves trigonometry, projectile motion formulas, and kinematic equations. Suitable for students in Grades 11-12.

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