Projectile Motion Lecture Notes

Problem Overview

• A ball rolls off an 80-meter cliff and lands 120 meters away from the base of the cliff.
• Objective: Calculate the initial velocity of the ball.

Key Information

• Height of the cliff (h): 80 meters
• Horizontal distance (range, R): 120 meters

Formula for Initial Velocity

• Initial Velocity (V₀) = ( R \sqrt{\frac{g}{2h}} )
  • Where:
    • R = Range
    • g = Gravitational acceleration (9.8 m/s²)
    • h = Height

Derivation of the Formula

1. Vertical Displacement Equation:
   • ( y_{final} = y_{initial} + V_{y}\cdot t - \frac{1}{2}gt^2 )
   • When y_initial = h and y_final = 0, it simplifies to ( h = \frac{1}{2}gt^2 )
   • Solve for t:
     • ( t = \sqrt{\frac{2h}{g}} )
2. Horizontal Displacement Equation:
   • ( R = V_{x}t )
   • With ( V_{x} = V_{0} ), we get ( V_{0} = \frac{R}{t} )
3. Combining these equations:
   • Substitute ( t ) into the initial velocity formula:
   • Resulting in ( V_{0} = R \sqrt{\frac{g}{2h}} )

Calculation

• Substitute values:
  • ( V_{0} = 120 \sqrt{\frac{9.8}{2 \times 80}} \approx 29.7 \text{ m/s} )_

Second Problem Overview

• A water hose shoots water from the ground at a 30-degree angle, landing 8 meters away.
• Objective: Calculate the initial velocity of the water.

Key Information

• Range (R): 8 meters
• Angle (θ): 30 degrees

Formula for Initial Velocity

• Initial Velocity (V₀) = ( \sqrt{\frac{Rg}{\sin(2\theta)}} )

Derivation of the Formula

1. Range Equation:
   • ( R = V_{x}t )
2. Time of Flight Equation:
   • ( t = \frac{2V_{y}}{g} )
3. Combine and Rearrange to derive the formula using the double angle formula for sine:
   • ( \sin(2\theta) = 2 \sin(θ) \cos(θ) )

Calculation

• Substitute values:
  • ( V_{0} = \sqrt{\frac{8 \times 9.8}{\sin(60)}} \approx 9.51 \text{ m/s at 30 degrees} )_

Third Problem Overview

• A ball launched at an angle from a 100-meter tall building lands 300 meters away after 10 seconds.
• Objective: Determine the initial launch angle and initial velocity.

Key Information

• Height (h): 100 meters
• Range (R): 300 meters
• Time (t): 10 seconds

Formula for Launch Angle

• Launch Angle (θ) = ( \tan^{-1}\left(\frac{\frac{1}{2}gt^2 - y_{initial}}{R}\right) )_

Derivation of the Formula

• Use vertical displacement and horizontal displacement equations similar to previous problems.

Calculation for Launch Angle

• Substitute values:
  • ( θ = \tan^{-1}\left(\frac{\frac{1}{2} \times 9.8 \times 100 - 100}{300}\right) \approx 52.43 \text{ degrees} )

Calculation for Initial Velocity

• Use the rearranged formula:
  • ( V_{0} = \frac{R}{t \cdot \cos(θ)} )
• Substitute values:
  • ( V_{0} = \frac{300}{10 \cdot \cos(52.43)} \approx 49.2 \text{ m/s} )

Summary of Results

• Problem 1: Initial Velocity = 29.7 m/s (0 degrees)
• Problem 2: Initial Velocity = 9.51 m/s (30 degrees)
• Problem 3: Launch Angle = 52.43 degrees, Initial Velocity = 49.2 m/s

Additional Resources

• More problems can be found in the provided links.
• Video tutorial website: video-tutor.net