Projectile Motion and Complementary Angles

Homework Discussion and Projectile Motion Concepts

Initial Conditions and Complementary Angles

• Initial velocity v = 25 m/s (same for all cases discussed).

• Two launch angles: 35° and 55°. These are complementary angles (35° + 55° = 90°).

  • Complementary angles result in the same range when initial velocity is constant and the projectile lands at the same elevation from which it was launched.

• Initial vertical velocity component for 35°: v_{y} = 25 \sin(35°) = 14.34 m/s

• Vertical velocity component for 55°: v_{y} = 25 \sin(55°) = 20.48 m/s

Complementary Angles and Trigonometric Functions

• Sine of an angle equals the cosine of its complement:

\sin(\theta) = \cos(90° - \theta)

Problem: Max Height, Range, and Hang Time

• The task is to calculate the max height, range, and hang time for both launch angles (35° and 55°) with an initial velocity of 25 m/s.

Results

• Range: The range is the same for both angles because they are complementary.

  • Range is calculated using the formula: R = \frac{v^2 \sin(2\theta)}{g}, where v is initial velocity, \theta is the launch angle, and g is the acceleration due to gravity.

• Max Height: The smaller angle (35°) has a lower max height, approximately half of the max height of the larger angle (55°).

  • Max height is calculated using: H = \frac{(v \sin(\theta))^2}{2g}, where v is initial velocity, \theta is the launch angle, and g is the acceleration due to gravity.

• Hang Time: The larger angle (55°) has a greater hang time than the smaller angle (35°).

  • Hang time is calculated using: T = \frac{2v \sin(\theta)}{g}, where v is initial velocity, \theta is the launch angle, and g is the acceleration due to gravity.

Proven Concept

• Objects launched with the same initial velocity at complementary angles travel the same range, assuming they return to the same elevation as the launch point.

• This was demonstrated with angles 35° and 55°, and implied for 30° and 60° from a previous assignment.

Bonus Assignment (2 points)

1. Algebraic Proof: Prove algebraically that objects launched at complementary angles with the same initial velocity will always have the same range.

   • Use the relationship: \sin(\theta) = \cos(90° - \theta)

   • Solve the range equation using variables (\theta and 90° - \theta) instead of numbers and demonstrate that the range is equal for both.

2. Optimum Angle for Maximum Range: Prove why a 45-degree launch angle maximizes the range.

   • Hint: Use calculus to find the maximum value of the range equation with respect to the launch angle.

Maximizing Range

• Prediction: A 45-degree angle will maximize the range.

• Reasoning: At 45 degrees, the sine and cosine components are equal, potentially maximizing the product that determines range.

• Implications:

  • Sports Applications: Throwing (shot put), jumping (long jump), softball (throwing from the outfield).

  • The optimal launch angle to maximize range is 45 degrees in these scenarios.

  • Trajectory: Discuss the effects of launch angle on trajectory shape and range.

Homework Review: Problems 4, 5, and 6

• Problems 4 and 5 were assigned for whiteboard presentation, while problem 6 was assigned to the rest of the class.

Key Considerations for Problem Solving

1. Organizing Variables: Use a chart format with x and y components.

   • Example: List initial and final positions, velocities, accelerations, and times in both x and y directions.

2. Subscripts: Use subscripts to represent the time interval (e.g., position 1 to position 2) for each variable (velocities, positions) being used.

3. Choosing positions: position one, the launch point. Position 2 is right before it lands.

• When finding max height, consider that you can only use variables appropriate from one position to the other. If, for example, you are looking for max height (displacement in y direction from Position 1 to Position 2), you can only use the variables associated with those positions. So, if you’re doing vf=at+vi, vi would be the initial velocity, vf would be the final velocity, and t will be your time interval.

Alternative Problem-Solving Approaches

• Instead of defining position 1 as the launch point and position 2 as right before impact, alternative positions can be chosen (e.g., max height as an intermediate position).

• Breaking the problem into segments:

  • Calculate parameters from launch to max height (where v_{y} = 0).

  • Then, calculate parameters from max height to the final position.

  • Discuss segmenting trajectory and its advantages in problem-solving.

Solving with Energy Considerations

1. Energy Bar Graphs: Construct energy bar graphs for different positions (e.g., position 1 for launch, position 1a, position 2 for impact).

2. Energy Components: Account for kinetic and gravitational potential energy at each point.

   • Position 1: Kinetic and gravitational energy.

   • Position 1a: Kinetic energy is the total kinetic energy in the system.

   • Describe the transformations between kinetic and potential energy during projectile motion.

3. Energy Conservation: Energy is conserved during the projectile motion (assuming no air resistance).

4. Finding final velocity: You would have to do some more work to find the velocity at position two would be one half mv_x^2

Buggy Lab and Graphing

1. Video Analysis: Use a video analysis tool to track the motion of a buggy moving in a circle.

2. Origin Setting: Set the origin at the center of the circular path to obtain symmetrical graphs.

3. Data Collection: Collect data to create six graphs.

   • Discuss common errors and ways to reduce these during data collection.

4. Resultant Velocity: Calculate the resultant velocity using the Pythagorean theorem from the velocity components (vx and vy). Rescale the graph to represent as a linear fit.

5. Linear Fit: Perform a linear fit on the resultant velocity graph. The slope should be approximately horizontal line indicating constant velocity.
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