A cannon with a muzzle speed of 996 m/s is used to start an avalanche on a mountain slope. The target is 2050 m from the cannon horizontally and 803 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

To solve this problem, we need to find the angle at which the cannon should be fired so that it hits a target at a specific horizontal distance and height. We'll use the following equations of projectile motion:

1. **Horizontal displacement**:
\[
x = v_0 \cdot \cos(\theta) \cdot t
\]

2. **Vertical displacement**:
\[
y = v_0 \cdot \sin(\theta) \cdot t - \frac{1}{2} g t^2
\]

Where:
- \( v_0 = 996 \, \text{m/s} \) (muzzle speed),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( x = 2050 \, \text{m} \) (horizontal distance to target),
- \( y = 803 \, \text{m} \) (vertical distance to target),
- \( \theta \) is the angle of projection,
- \( t \) is the time of flight.

### Step 1: Express Time \( t \)
From the horizontal displacement equation:
\[
t = \frac{x}{v_0 \cdot \cos(\theta)}
\]

### Step 2: Substitute \( t \) into the vertical displacement equation
Substitute \( t \) from the first equation into the vertical equation:
\[
y = v_0 \cdot \sin(\theta) \cdot \left(\frac{x}{v_0 \cdot \cos(\theta)}\right) - \frac{1}{2} g \left(\frac{x}{v_0 \cdot \cos(\theta)}\right)^2
\]
Simplifying:
\[
y = x \cdot \tan(\theta) - \frac{g x^2}{2 v_0^2 \cdot \cos^2(\theta)}
\]

### Step 3: Solve for \( \theta \)
We can now solve this equation numerically to find the value of \( \theta \). Let's calculate it.The angle at which the cannon should be fired is approximately \( 21.97^\circ \) above the horizontal.

Would you like more details on the steps involved or have any further questions?

Here are five related questions to consider:
1. How does air resistance affect projectile motion in real-world scenarios?
2. What is the total time of flight for this particular projectile?
3. How does varying the muzzle speed affect the required angle for hitting the target?
4. Can this method be used to solve for angles in other types of projectile motion problems?
5. What would happen if the target were lower instead of higher than the cannon?

**Tip**: In projectile motion problems, splitting the motion into horizontal and vertical components helps to simplify complex calculations.

A cannon with a muzzle speed of 996 m/s is used to start an avalanche on a mountain slope. The target is 2050 m from the cannon horizontally and 803 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

Learn how to calculate the firing angle of a cannon with a muzzle speed of 996 m/s to hit a target 2050 m away horizontally and 803 m above the cannon. This detailed solution covers projectile motion, using key kinematic equations and trigonometric formulas. Suitable for Grades 11-12.

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