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Part 1
A projectile fired from the point​ (0,0) at an angle to the positive​ x-axis has a trajectory given by y equals Cx minus left parenthesis 1 plus Upper C squared right parenthesis left parenthesis StartFraction g Over 2 EndFraction right parenthesis left parenthesis StartFraction x Over v EndFraction right parenthesis squared .
In this​ formula, x is the horizontal distance in​ meters, y is the height in​ meters, v is the initial velocity in meters per​ second, gequals9.81 m divided by sec squared is the acceleration due to​ gravity, and Upper C greater than 0 is a constant determined by the angle of elevation.
A howitzer fires an artillery round with a velocity of 883 ​m/sec. Answer parts​ (a) and​ (b).
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Part 1
​(a) If the round must clear a hill 156 meters high at a distance of 1946 meters in front of the​ howitzer, what C values are permitted in the trajectory​ equation?

To solve for the permissible values of \( C \) in the trajectory equation, we need to use the given trajectory equation:

\[
y = Cx - \left( 1 + C^2 \right) \left( \frac{g}{2} \right) \left( \frac{x}{v} \right)^2
\]

where:

- \( x \) is the horizontal distance,
- \( y \) is the height,
- \( v \) is the initial velocity,
- \( g = 9.81 \, \text{m/s}^2 \) (the acceleration due to gravity),
- \( C \) is a constant related to the angle of elevation.

We are given:
- The hill height \( y = 156 \, \text{m} \),
- The distance \( x = 1946 \, \text{m} \),
- The initial velocity \( v = 883 \, \text{m/s} \).

Now, substitute the known values into the trajectory equation to find \( C \):

\[
156 = C(1946) - \left( 1 + C^2 \right) \left( \frac{9.81}{2} \right) \left( \frac{1946}{883} \right)^2
\]

We'll solve this equation for \( C \).

### Step-by-Step Solution

1. Calculate \( \left( \frac{1946}{883} \right)^2 \).
2. Substitute that value and simplify the equation.
3. Solve the resulting quadratic equation for \( C \).

Let me calculate that now.The permissible values for \( C \) in the trajectory equation are approximately:

- \( C \approx 0.093 \)
- \( C \approx 81.592 \)

These two values correspond to different angles of elevation for the howitzer, one with a much smaller angle and the other with a much larger angle, both allowing the projectile to clear the hill at 156 meters height and 1946 meters distance.

Would you like to know more details on how these values affect the trajectory, or have any other questions?

Here are 5 related questions:
1. How do different values of \( C \) affect the overall trajectory of the projectile?
2. Can you determine the angle of elevation from the value of \( C \)?
3. How would air resistance alter the equation and the values of \( C \)?
4. What happens if the initial velocity is increased or decreased?
5. How would changing the height or distance of the hill affect the solution?

**Tip:** When solving projectile motion problems, always ensure that units are consistent to avoid calculation errors.

A projectile fired from the point (0,0) at an angle to the positive x-axis has a trajectory given by y = Cx - (1 + C^2)(g/2)(x/v)^2. In this formula, x is the horizontal distance in meters, y is the height in meters, v is the initial velocity in meters per second, g = 9.81 m/s^2 is the acceleration due to gravity, and C > 0 is a constant determined by the angle of elevation. A howitzer fires an artillery round with a velocity of 883 m/s. (a) If the round must clear a hill 156 meters high at a distance of 1946 meters in front of the howitzer, what C values are permitted in the trajectory equation?

Learn how to solve for the constant C in a projectile motion problem involving a howitzer firing an artillery round with an initial velocity of 883 m/s. The round must clear a hill 156 meters high at a distance of 1946 meters. This problem covers key kinematic concepts and quadratic equations, suitable for high school students in Grades 11-12.

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