The trajectory equation

Eliminate time: y = x tanθ − gx²/(2u²cos²θ) — the path is a parabola.

The trajectory equation0 min · Free lecture

In this lesson

  • Eliminate time: y = x tanθ − gx²/(2u²cos²θ) — the path is a parabola.

Every projectile follows the same shape: a parabola, whatever the launch data.

Eliminate t between x and y to get the trajectory. For a given launch, the path is fixed; the equation answers 'where is the projectile at this x?' without any time bookkeeping.

The coefficient of x² is −g/(2u²cos²θ) — flatter launch, shallower parabola.

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}

Trajectory of a projectile

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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The trajectory equation — FemtoLearn.
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Eliminate time: y = x tanθ − gx²/(2u²cos²θ) — the path is a parabola.
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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°