Projectile graphs

y(t) is a parabola; v_y(t) is a straight line; x(t) is linear — recognise the family.

Projectile graphs0 min · Free lecture

In this lesson

  • y(t) is a parabola; v_y(t) is a straight line; x(t) is linear — recognise the family.

Three graphs, three shapes: line, line, parabola.

x(t): straight line (slope u cosθ). v_y(t): straight line (slope −g). y(t): inverted parabola with peak at T/2. a_y(t): horizontal line at −g.

Graph questions usually test one recognition: which of these is NOT a projectile graph, or which pair belongs together.

x(t)=ucos⁡θ t,y(t)=usin⁡θ t−12gt2x(t) = u\cos\theta\, t,\quad y(t) = u\sin\theta\, t - \tfrac12 gt^2

Projectile coordinates as functions of time

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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Projectile graphs — FemtoLearn.
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y(t) is a parabola; v_y(t) is a straight line; x(t) is linear — recognise the family.
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°