The maximum-range angle (45°)

At 45° the range peaks; complementary angles share the same range.

The maximum-range angle (45°)0 min · Free lecture

In this lesson

  • At 45° the range peaks; complementary angles share the same range.

45° is the sweet spot: equal horizontal and vertical components.

R ∝ sin2θ peaks at 2θ = 90°, i.e. θ = 45°. For any θ, the angle 90°−θ lands at the same distance — with a shorter flight time (θ > 45° flies longer).

On an incline, the optimal angle shifts to (45° − α/2) for uphill projection — the advanced twist.

sin⁡2θ=1⇒θ=45∘\sin 2\theta = 1 \Rightarrow \theta = 45^\circ

Condition for maximum range

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°

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The maximum-range angle (45°) — FemtoLearn.
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At 45° the range peaks; complementary angles share the same range.
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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°