Horizontal range

R = u²sin2θ/g — the horizontal velocity times the flight time.

Horizontal range0 min · Free lecture

In this lesson

  • R = u²sin2θ/g — the horizontal velocity times the flight time.

Range is a product of horizontal speed and flight time — both change with θ.

R = uₓ·T = u cosθ × 2u sinθ/g = u²sin2θ/g.

Two complementary angles (θ and 90°−θ) give the same range — a favourite JEE fact. Maximum range at 45°: R_max = u²/g.

R=u2sin⁡2θg,Rmax=u2gR = \frac{u^2\sin 2\theta}{g},\quad R_{max} = \frac{u^2}{g}

Range and 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°

Transcript (0 min)
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Horizontal range — FemtoLearn.
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R = u²sin2θ/g — the horizontal velocity times the flight time.

Frequently asked

Which angle gives the maximum range?

45°, giving R = u²/g.

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 · 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°
  3. Q3 · 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}
  4. Q4 · 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.