Relative motion of two projectiles

The relative acceleration between two projectiles is zero — they move in straight lines relative to each other.

Relative motion of two projectiles0 min · Free lecture

In this lesson

  • The relative acceleration between two projectiles is zero — they move in straight lines relative to each other.

Two projectiles launched together have constant relative velocity: straight-line relative paths.

a_rel = a₁ − a₂ = (−g) − (−g) = 0. So the relative motion is uniform: r_rel(t) = r₀ + v_rel t — a straight line.

Collision checks between projectiles become linear algebra: does r₁(t) = r₂(t) have a positive time solution?

r⃗rel(t)=r⃗0+v⃗rel t\vec{r}_{rel}(t) = \vec{r}_0 + \vec{v}_{rel}\, t

Relative position of projectiles

Worked example

For a fixed launch speed, the horizontal range of a projectile is maximum when the launch angle is:

  1. R = v0² sin 2θ / g
  2. R is maximum when sin 2θ = 1, i.e. 2θ = 90°, so θ = 45°

Answer: 45°

Transcript (0 min)
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Relative motion of two projectiles — FemtoLearn.
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The relative acceleration between two projectiles is zero — they move in straight lines relative to each other.
Printable notes (free)

Practice

Free · 4 questions with full solutions
  1. Q1 · Numerical · difficulty 2/5

    A projectile is launched with speed 15 m/s at an angle 70° above the horizontal. Taking g = 10 m/s², find its horizontal range.

    R=v02sin⁡2θgR = \frac{v_0^2 \sin 2\theta}{g}
  2. Q2 · Numerical · difficulty 2/5

    A projectile is launched with speed 15 m/s at an angle 25°. Taking g = 10 m/s², find its maximum height above the launch point.

    H=v02sin⁡2θ2gH = \frac{v_0^2 \sin^2\theta}{2g}
  3. Q3 · Numerical · difficulty 2/5

    A projectile is launched with speed 25 m/s at an angle 15°. Taking g = 9.8 m/s², find its total time of flight.

  4. Q4 · MCQ · difficulty 1/5

    For a fixed launch speed, the horizontal range of a projectile is maximum when the launch angle is:

    • 30°
    • 45°
    • 60°
    • 90°