Lec 04: The Motion of Projectiles | 8.01 Classical Mechanics, Fall 1999 (Walter Lewin)

Main ideas / concepts / lessons

• Projectile motion setup (2D motion with gravity)

  • A golf/tennis ball is launched from the ground with initial speed (v_0) at an angle (\alpha).
  • Coordinate choice:
    • Horizontal direction: increasing (x)
    • Vertical direction: increasing (y)
  • Initial velocity components:
    • (v_{0x} = v_0\cos\alpha)
    • (v_{0y} = v_0\sin\alpha)
  • Motion assumptions used:
    • No acceleration in (x) (so horizontal velocity stays constant)
    • Acceleration in (y) due to gravity: downward acceleration of magnitude (g) (taken as (9.8\ \text{m/s}^2))

• Using standard kinematic equations

  • Horizontal motion (1D):
    • (x(t) = v_0\cos\alpha \; t)
  • Vertical motion (1D):
    • (y(t) = v_0\sin\alpha \; t - \tfrac12 g t^2)

• Eliminating time to get the trajectory shape

  • Eliminate (t) using (t = \dfrac{x}{v_0\cos\alpha}).
  • Resulting trajectory:
    • (y(x)) becomes a parabola (second-order in (x))
  • Key takeaway: projectile trajectories are parabolic when air drag is neglected.

• Highest point of the projectile

  • Condition: at the top, vertical velocity is zero.
    • (v_y(t) = v_0\sin\alpha - gt)
    • Set (v_y=0) ⇒
      • (\displaystyle t_P = \frac{v_0\sin\alpha}{g})
  • Height above ground at the top:
    • (\displaystyle h = y(t_P) = \frac{v_0^2\sin^2\alpha}{2g})

• Time to land and total range

  • Time to reach the ground point (S):
    • Using symmetry of the parabola about the peak:
      • (\displaystyle t_S = 2t_P = \frac{2v_0\sin\alpha}{g})
  • Horizontal distance from launch to landing:
    • (OS = v_{0x}\,t_S = (v_0\cos\alpha)\left(\frac{2v_0\sin\alpha}{g}\right))
    • Using (2\sin\alpha\cos\alpha = \sin(2\alpha)):
      • (\displaystyle OS = \frac{v_0^2\sin(2\alpha)}{g})
  • Reasoning for why (v_0^2) appears (conceptual argument):
    • Doubling speed roughly doubles flight time, and also doubles horizontal velocity ⇒ range scales like (2\times 2 = 4), i.e., proportional to (v_0^2).

• Launch angle for maximum distance (in ideal vacuum model)

  • From (\sin(2\alpha)), maximum occurs at:
    • (2\alpha = 90^\circ \Rightarrow \alpha = 45^\circ)
  • Caveat:
    • Real sports involve air drag, spin, etc., so real optimal angles differ.

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Experimental validation and uncertainty handling (methodology)

• Goal of experiment

  • Test predicted landing distances for launch angles (given the same measured launch “power”/speed behavior).

• Major measured quantity needed

  • The lecturer focuses on (v_0^2) rather than (v_0).
  • Reason:
    • The range formula uses (v_0^2), so uncertainties in the vertical maximum height directly propagate to (v_0^2).

• Step-by-step procedure (as described)

  1. Estimate uncertainties in the measured initial speed
     • Launch vertically ((\alpha = 90^\circ)).
     • Vertical maximum height:
       • (\displaystyle h_{\max} = \frac{v_0^2}{2g})
     • Measure (h_{\max}) relative to a 3-meter stick top mark.
     • Repeat multiple trials (done “20 times”).
     • Determine a conservative height uncertainty:
       • Typically differences sometimes reached 10–15 cm
       • Lecturer chooses 15 cm uncertainty in (h_{\max})
  2. Convert height uncertainty to uncertainty in (v_0^2)
     • Since (v_0^2 = 2gh_{\max}), uncertainty in (v_0^2) inherits the same percentage uncertainty as (h_{\max}).
     • Using the measured value:
       • (h_{\max} \approx 3.07\ \text{m})
       • So:
         • (v_0^2 \approx 60.2\ (\text{m}^2/\text{s}^2))
       • Chosen uncertainty corresponds to about 5% in (v_0^2).
  3. Analyze angle uncertainty separately
     • Assume angle can only be set accurately to about ±1°.
     • For (\alpha=45^\circ):
       • (\sin(2\alpha) = \sin(90^\circ) = 1)
       • Even if the angle is off by 1°, (\sin(88^\circ)) remains extremely close to 1.
     • Result: angle uncertainty contributes negligibly compared to (v_0^2) uncertainty.
  4. Make range predictions using uncertainty propagation
     • Range formula:
       • (\displaystyle OS = \frac{v_0^2\sin(2\alpha)}{g})
     • For 45°:
       • Use (\sin(2\alpha)=1)
       • So:
         • (\displaystyle OS \approx \frac{60.2}{9.8} \approx 6.14\ \text{m})
       • With 5% uncertainty, expected landing band has width corresponding to ~30 cm (as stated).
  5. Perform trials at the predicted angles
     • Lecturer physically fires at 45° and points to where the ball lands relative to the marked uncertainty interval.
  6. Repeat reasoning for 30°
     • For (\alpha=30^\circ):
       • (\sin(2\alpha)=\sin(60^\circ)\approx 0.866)
       • With ±1° (so (\sin(58^\circ)) and (\sin(62^\circ))):
         • Relative change in (\sin(2\alpha)) becomes about 2%
     • Combine uncertainties:
       • ~5% from (v_0^2) plus ~2% from (\sin(2\alpha)) ⇒ overall ~7% uncertainty in range
       • This yields a larger uncertainty band (~37 cm stated) and a wider “allowed” hit region.
  7. Check symmetry at 60°
     • Range depends on (\sin(2\alpha)).
     • (\sin(120^\circ)=\sin(60^\circ)) ⇒ same predicted range for 30° vs 60° (with the same speed and uncertainty logic).

• Key experimental lesson

  • Including measurement uncertainties is crucial:
    • It explains why results can look “wrong” in high school level certainty terms, yet still be consistent with physics once error bounds are included.

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“Monkey and hunter” thought experiment (conceptual lesson)

• Tragic scenario

  • A hunter aims a golf ball at a monkey at some location.
  • The lecturer argues what happens when gravity affects both “no gravity” and “with gravity” trajectories.

• Core physics argument

  • Compare two hypothetical paths:
    • No gravity path (pure straight-line due to constant velocity components)
    • With gravity path (actual projectile parabola)
  • At any given time, the horizontal position (x) is the same with or without gravity because there is no horizontal acceleration.
  • The vertical separation between the trajectories at that same time is:
    • (\displaystyle \Delta y = \tfrac12 g t^2)
  • The time (t_2) when the ball’s horizontal position matches the monkey’s causes the monkey to have fallen the same vertical distance (\tfrac12 g t_2^2).
  • Therefore, under the “perfect aim” condition, the monkey is hit.

• Critical dependence

  • The outcome depends mainly on the aim angle (and thus trajectory alignment).
  • If the angle is off enough, the monkey might survive because the ball and monkey no longer coincide.

• Reference-frame check (elevator viewpoint)

  • The lecturer re-derives the “time to kill” in a free-fall (elevator) frame:
    • In free fall, the monkey sees the bullet approach in a straight line (no arc).
  • The monkey computes time using distance and speed:
    • Time (= \dfrac{\sqrt{d^2+h^2}}{v_0})
  • He also shows agreement with the ground-frame computation using horizontal speed:
    • (t_{\text{kill}} = \dfrac{D}{v_0\cos\alpha})
  • The final message: results are consistent across reference frames.

• Live demo

  • A “monkey” prop (“Robert”) is suspended and released (via electromagnet) while a golf ball is fired so it lands at the predicted intersection point.
  • The lecturer notes it is set up carefully and the prop is protected (bulletproof vest).

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Speakers / sources featured

• Walter Lewin (lecturer; speaker throughout)
• Students / CLASS (audience responses and participation during experiments)
• “MIT students t” / “MIT students” (referenced as the context; not individual named speakers)