The geometry of a launch

Why does a projectile trace a parabola?

Why does a projectile trace a parabola?

  1. Compare launches at 30° and 60°.
  2. Set 45°, then double launch speed.
  3. Reduce gravity and compare flight time and range.

What to measure

v₀ in m/s; g in m/s²; θ is launch angle; R in m.

The idea underneath

Horizontal motion is uniform while vertical motion has constant downward acceleration. Combining these two independent motions produces a parabola. At the apex vertical velocity is zero but acceleration remains −g. For launch and landing at equal heights, complementary angles give equal ranges, and 45° gives the greatest range only under the ideal no-drag assumptions. Each playback cycle repeats a fresh launch; it does not add energy to a bouncing object.

x=v₀ cosθ·t · y=v₀ sinθ·t−½gt² · R=v₀² sin(2θ)/g

v₀ in m/s; g in m/s²; θ is launch angle; R in m.

Build the argument

  1. Draw the forces and choose coordinates before writing a differential equation.
  2. Check the result against energy conservation and dimensional consistency.
  3. Change one parameter at a time and compare a ratio rather than relying on visual impressions.

Worked example 1

v₀=8 m/s, θ=45°, g=9.81 gives R=6.524 m.

Worked example 2

The apex is 1.631 m and flight time is 1.153 s.

Applications

Oscillators appear in clocks, vibration isolation, sensors and mechanical structures. Conserved quantities make complicated motion easier to predict.

Common misconception

A force changes velocity, not just speed. A massless rod or ideal spring is a model assumption, and small-angle pendulum formulas have a limited range.

A further investigation

Find two different angles with the same range.

Check your understanding

At the apex, acceleration is…

  • Downward, g
  • Zero
  • Horizontal
Explanation

Gravity acts throughout the flight. Complementary angles have equal sin(2θ).

Model limits

Point projectile, flat ground, constant g, no drag, equal launch and landing heights.

OpenStax · Projectile motion