Projectile on an inclined plane

Project up an incline: rotate the axes — g splits into g sinα along and g cosα normal to the plane.

Projectile on an inclined plane0 min · Free lecture

In this lesson

  • Project up an incline: rotate the axes — g splits into g sinα along and g cosα normal to the plane.

An incline is a projectile problem in a tilted frame.

With x along the incline and y normal to it, the effective gravity has components g sinα and g cosα. The range along the incline: R = 2u²cosθ·sin(θ−α)/(g cos²α) — from the standard derivation.

The optimal projection angle up an incline is θ = 45° + α/2.

Rincline=2u2cos⁡θsin⁡(θ−α)gcos⁡2αR_{incline} = \frac{2u^2\cos\theta\sin(\theta-\alpha)}{g\cos^2\alpha}

Range along an incline

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 on an inclined plane — FemtoLearn.
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Project up an incline: rotate the axes — g splits into g sinα along and g cosα normal to the plane.
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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°