Projection from a height at an angle

From height h at angle θ: solve y = h with the full equations — the quadratic gives two times.

Projection from a height at an angle0 min · Free lecture

In this lesson

  • From height h at angle θ: solve y = h with the full equations — the quadratic gives two times.

Thrown from a cliff, the projectile lands below the launch level — the landing height is −h.

With the launch point as origin and down negative: y(t) = u sinθ t − ½gt² and the ground is y = −h. Set y = −h and solve; the positive root is the landing time.

The negative root is the ghost time when the projectile 'would have been' at that level before launch — discard it.

−h=usin⁡θ t−12gt2-h = u\sin\theta\, t - \tfrac{1}{2}gt^2

Landing condition from a height

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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Projection from a height at an angle — FemtoLearn.
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From height h at angle θ: solve y = h with the full equations — the quadratic gives two times.
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°