Projectile landing on an incline

Landing on a slope: the ground is y = −x tanα — intersect it with the trajectory.

Projectile landing on an incline0 min · Free lecture

In this lesson

  • Landing on a slope: the ground is y = −x tanα — intersect it with the trajectory.

An incline is just a slanted ground: y = −x tanα, and the trajectory meets it.

Set the trajectory equation equal to the incline line; the intersection gives the landing x along the slope, then convert to range along the incline.

Same physics as 'projectile on an incline' but viewed in the ground frame — pick whichever frame makes the algebra cleaner.

y=−xtan⁡αy = -x\tan\alpha

The incline as a line

Worked example

A projectile is fired up a plane inclined at 30° to the horizontal with speed u at 60° to the horizontal. Find its range along the incline (in terms of u and g).

  1. Use the incline range formula with θ = 60°, α = 30°.
  2. R = 2u²cosθ·sin(θ−α)/(g cos²α) = 2u²cos60°·sin30°/(g cos²30°).
  3. R = 2u²(½)(½)/(g·¾) = 2u²/(3g).

Answer: R = 2u²/(3g).

Transcript (0 min)
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Projectile landing on an incline — FemtoLearn.
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Landing on a slope: the ground is y = −x tanα — intersect it with the trajectory.
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°