Study Notes on Linear Motion Part 2: Free Fall

LMMU Levy - Mwanawasa University

LECTURE 4: PHY 101 - LINEAR MOTION PART 2: FREE FALL

Instructor: CHONGO, E.

Course Material

Introduction to Free Fall

• Definition:

  • Free fall describes the motion of an object under the influence of gravitational force only, occurring within a gravitational field like that near Earth's surface.

• Significance:

  • An important application of the equations of uniform motion.

  • Governs the motion under constant acceleration provided by gravity.

Characteristics of Free Fall

• Constant Acceleration:

  • Acceleration remains consistent due to Earth's gravity.

  • It is a critical case of constant acceleration.

• Independence of Mass:

  • The acceleration due to gravity does not depend on the mass of the object.

• Magnitude of Free-Fall Acceleration:

  • Denote by:

  • |a| = g = 9.81 \, m/s^2

• Direction of Acceleration:

  • Direction is always downward.

  • Can be represented as:

  • a = -g = -9.81 \, m/s^2 (if 'up' is defined as the positive direction).

Misconceptions Around Free Fall

• Clarification:

  • The term 'free fall' does not imply that objects can only fall downwards.

  • Examples include:

  • Objects thrown upwards

  • Objects released from rest

  • Objects falling downwards

One-Dimensional Motion Involving Gravity

• Assumptions:

  • For motion under gravity, assume acceleration is -g (positive direction defined as upward).

Key Equations

1. Final velocity equation:
   V_{yf} = V_{yi} - gt

2. Vertical position equation:
   y_f = y_i + V_{yi}t - rac{1}{2}gt^2

Detailed Examples

Example 1: Coin Flip

• Situation: You flip a coin straight up to decide on homework.

• Questions:
  a. What is the velocity of the coin at the top of its trajectory?
  b. If the coin reaches a height of 0.25 m, what was its initial speed?
  c. How long did it spend in the air?

• Solutions: a. At the top:

  • V_{top} = 0 \, m/s;

  • Acceleration: a = 9.8 \, m/s^2 downward.
    b. Using:

  • V_{y1}^2 = V_{y1}^2 + 2g ext{x} where V_{y1} = 0 at the height of the toss.

  • Solve for V_{y1}:
    V_{y1} = ext{√}(2gAx)
    = ext{√}(2)(9.8 \, m/s^2)(0.25 \, m) = 2.2 \, m/s
    c. Using:

  • V_{y1} = V_{i} + at

  • Rearranging gives:
    t = rac{V_{y1} - V_{yi}}{-g}

  • Plugging in:
    t = rac{-2.2 \, m/s - 2.2 \, m/s}{-9.8 \, m/s^2} = 0.45 \, s

Example 2: Upward Thrown Stone

• Situation: A stone thrown upwards with a speed of 20 m/s, caught 5.0 m above the throw point.

• Questions:
  a. How fast was it going when caught?
  b. How long did the trip take?

• Solutions:

  1. Given: v_{yi} = 20 m/s, y = 5.0 m.

  2. Using:

  • v^2_f = v^2_{yi} - 2g(y_f - y_i)

  • Thus,
    v^2_f = (20 \, m/s)^2 - 2(9.81 \, m/s^2)(5.0 \, m - 0)

  • Solving gives:
    v_f = ext{±} ext{√}(302) = -17.4 \, m/s

  • Note: The negative indicates the stone is descending.

  1. Time calculation using:
     V_{f} = V_{i} - gt

  • Rearranging gives:
    t = rac{V_{f} - V_{i}}{-g}

  • Plugging in:
    t = rac{-17.4 \, m/s - 20 \, m/s}{-9.81 \, m/s^2} = 3.8 \, s

Example 3: Brick Drop

• Situation: A brick dropped from a high scaffold.

• Questions:
  a. What is the velocity after 4.0 s?
  b. How far does the brick fall?

• Solutions: a. Using the formula:

  • V_f = V_i + at

  • Here, V_i = 0 \, m/s and a = 9.8 \, m/s^2.

  • Thus:
    V_f = 0.0 \, m/s + (-9.8 \, m/s^2)(4.0 \, s) = -39.2 \, m/s

  • Direction is downward, so speed is 39.2 m/s downward.
    b. Using:

  • x = V_it + rac{1}{2}at^2

  • Therefore,
    x = 0 + rac{1}{2}(-9.81 \, m/s^2)(4.0 \, s)^2

  • Resulting in:
    x = -78.4 \, m (the brick falls approximately 78 m).

Task

• Situation: A ball thrown vertically upward on the moon returns to its starting point in 4.0 s under a gravity of 1.60 \, m/s^2.

• Question: What was the ball's original speed?

• Answer:

  • Original speed: V_{yi} = 3.2 \, m/s