Free Fall Example Problems

Main ideas / lessons

• These are worked free-fall (kinematics) example problems using standard equations under constant gravitational acceleration.
• The key gravitational acceleration magnitude is always 9.80 m/s², but the sign (+/−) depends on whether the object’s motion is defined as upward positive or downward positive (or, as stated, depending on whether the problem is a “downward fall” or an “up-down” situation).
• For free-fall problems:
  • Acceleration is fixed at 9.80 m/s² (with sign chosen appropriately).
  • Distance (height) is treated as a position change along the vertical direction.
  • Common strategy: use the kinematics equation that matches exactly the known variables.
• Dropped from rest is treated as initial velocity = 0 m/s (even if not explicitly stated).
• For an object thrown straight upward:
  • At the top, final velocity is 0 m/s.
  • The time to go up equals the time to come back down, so total time = 2 × (time to top).

Methodology / instruction list (as presented)

When solving a free-fall problem:

• Identify whether it’s:
  • Downward-only (dropped straight down), or
  • Up-and-down (thrown upward).
• Write down acceleration:
  • Use +9.80 m/s² for a downward/falling definition where positive direction matches the fall.
  • Use −9.80 m/s² for an up/down setup where upward is treated positive and gravity opposes upward motion.
• Convert/assign known quantities:
  • Height given in meters is distance traveled in the vertical direction.
  • If “dropped” from rest: set (v_i = 0).
• Note variable availability:
  • Use kinematics equations when you have the needed set of variables.

• Pick the appropriate kinematics equation:

  • Equation 3 (distance form) used when solving for time with given distance, acceleration, and initial velocity: [ d = v_i t + \tfrac{1}{2} a t^2 ]

  • Equation 4 used when solving for height using (v_f^2): [ v_f^2 = v_i^2 + 2ad ]

  • Equation 1 used for time with known velocities and acceleration: [ v_f = v_i + at ]

• Solve algebraically, then:

  • When using an equation containing (t^2), take the square root to isolate (t).
• For significant figures:
  • Keep appropriate rounding (stated: 3 sig figs based on given numbers like 9.80 and 15.0).

For “thrown upward” problems:

• Set (v_f = 0) at the top to simplify finding maximum height.
• Compute time to top, then:
  • Total time in air = 2 × (time to top).

Worked example outcomes

Free-fall example #1 (dropped downward)

• Given: a ball dropped from 15.0 m, starting from rest.
• Acceleration: +9.80 m/s² (direction chosen consistent with downward motion).
• Initial velocity: (v_i = 0).
• Unknown: time to hit ground.

• Using: [ d = v_i t + \tfrac{1}{2} a t^2 ] [ 15 = 0 + 4.90t^2 ] [ t = \sqrt{15/4.90} \approx 1.75 \text{ s} ]

• Answer: 1.75 s

Free-fall example #2 (thrown upward)

• Given: ball thrown straight upward at 13 m/s.
• Acceleration: −9.80 m/s² (gravity opposes upward motion).
• Initial velocity: (v_i = +13 \text{ m/s}).

Part (a) Maximum height

• At top: choose (v_f = 0).

• Use: [ v_f^2 = v_i^2 + 2ad ] [ 0 = 13^2 + 2(-9.80)d ] [ 0 = 169 - 19.6d ] [ d \approx 8.62 \text{ m} ]

• Answer: 8.62 m (height reached)

Part (b) Total time in air

• First compute time to top using: [ v_f = v_i + at ] [ 0 = 13 + (-9.80)t ] [ t = 13/9.80 \approx 1.33 \text{ s} ]

• Total time: twice the time to top: [ T = 2(1.33) \approx 2.66 \text{ s} ]

• Answer: 2.66 s

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