Relations among R, H and T

R = 4H cotθ, T² = 2H/g·(4/... ) — the three results interlock; one given value fixes the rest.

Relations among R, H and T0 min · Free lecture

In this lesson

  • R = 4H cotθ, T² = 2H/g·(4/... ) — the three results interlock; one given value fixes the rest.

Know any two of R, H, T and the launch is fully determined.

Key identities: H = u²sin²θ/2g, R = u²sin2θ/g, T = 2u sinθ/g give R = 4H cotθ and T² = 8H/g.

These let you jump between flight data without recovering u and θ — fast paths in multi-part questions.

R=4Hcot⁡θR = 4H\cot\theta

Range in terms of height and angle

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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Relations among R, H and T — FemtoLearn.
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R = 4H cotθ, T² = 2H/g·(4/... ) — the three results interlock; one given value fixes the rest.
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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°