Projectile Motion

Introduction to Projectile Motion

• Objectives:

  • Sketch the theoretical path of a projectile.

  • Understand the independence of vertical and horizontal motion.

  • Solve problems involving projectiles launched horizontally and at angles.

Definition of a Projectile

• A projectile is defined as an object with only gravity acting upon it (ignoring air resistance).

• Example: A baseball thrown in the air, primarily influenced by gravity.

Key Characteristics of Projectile Motion

• Neglect Air Resistance:

  • Focus on a simplified model; real-world scenarios may differ.

• Projectile Path:

  • Typically follows a parabolic arc, especially when launched at an angle from the same height.

  • The time to ascend equals the time to descend.

  • At equal heights: Initial and final velocity magnitudes are the same but in opposite directions.

Independence of Motion

• Vertical and horizontal motions are independent:

  • Vertical motion affected by gravity, utilizing free-fall equations.

  • Horizontal motion remains constant, with no acceleration.

  • Final horizontal velocity = Initial horizontal velocity.

• Key Concept: The time the object is in the air is the same for both vertical and horizontal motion, enabling the use of time from vertical calculations for horizontal calculations.

Important Notes on Projectile Behavior

• An object maintains projectile motion only while gravity is the sole force acting on it.

  • Example: A thrown ball is a projectile until caught or it contacts the ground.

Problem-Solving Example

Problem Statement

• Fred throws a baseball horizontally at 42 m/s from a height of 2 m. Find horizontal distance traveled before it hits the ground.

Solution Steps

• Horizontal Motion Table:

  • Initial Horizontal Velocity, $V_x = 42 ext{ m/s}$

  • Acceleration, $A = 0$

  • Projectile Time (T) will be derived from vertical motion.

Vertical Motion Calculation

• With initial vertical velocity $V_Y = 0$, and displacement Delta Y = 2, the equation simplifies to:

• Delta Y = √ (9.8 ext{ m/s}^2) T²

• Solve for T

Using Time to Find Horizontal Range

• For horizontal motion:

  • $Delta X = V_{avg} imes T$

Additional Example: Stunt Car Problem

• A stunt car driven off a cliff will maintain its horizontal velocity due to the lack of horizontal forces.

Important Concept

• The horizontal component of velocity remains unchanged, whereas vertical overall velocity changes throughout the fall.

Launch Angle Considerations

Maximum Range

• The angle that provides maximum horizontal distance for projectile motion is 45° when launched from the same height.

Complex Problem with Components

Problem Statement

• Herman is launched at an angle of 30° with an initial velocity of 26 m/s.

Solution Steps

• Component Breakdown:

  • Horizontal Component:
    Delta V_x = 26 m/s{cos}(30°)
    = 22.5 m/s

  • Vertical Component:
    Delta V_y = 26 m/s{sin}(30°)
    = 13 m/s

Time Calculation

• Total time in the air = upward time × 2

• Vertical table setup:

  • For vertical motion, using not final height calculations:
    Delta T_up = √(0 - 13)} / {-9.8}
    = 1.33 ext{ seconds up, therefore } 2.65 ext{ seconds total}

Horizontal Motion for Distance Calculation

• Delta X = V_x T = 22.5({ m/s }) * 2.65 { seconds} = 59.6 m