Horizontal projection from a height

Thrown horizontally at u from height h: T = √(2h/g), R = u√(2h/g) — vertical time, horizontal distance.

Horizontal projection from a height0 min · Free lecture

In this lesson

  • Thrown horizontally at u from height h: T = √(2h/g), R = u√(2h/g) — vertical time, horizontal distance.

The horizontal throw never changes the fall time — it only adds distance.

Vertical: u_y = 0, so the fall time T = √(2h/g) is identical to a drop. Horizontal: x = u·T = u√(2h/g).

The impact speed: v = √(u² + (gT)²) — combine components at the end, never the averages.

T=2hg,R=u2hgT = \sqrt{\frac{2h}{g}},\quad R = u\sqrt{\frac{2h}{g}}

Horizontal projection

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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Horizontal projection from a height — FemtoLearn.
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Thrown horizontally at u from height h: T = √(2h/g), R = u√(2h/g) — vertical time, horizontal distance.
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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 · 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.