Projectile from a moving platform

Released from a moving balloon/plane: the projectile inherits the platform's velocity.

Projectile from a moving platform0 min · Free lecture

In this lesson

  • Released from a moving balloon/plane: the projectile inherits the platform's velocity.

A bomb dropped from a plane flies forward with the plane's speed — it doesn't fall straight down.

On release, the object keeps the platform's velocity components. From a plane flying level: uₓ = plane speed, u_y = 0 — a horizontal projection from the release height.

The horizontal distance to the target: R = u√(2h/g); the pilot must release early by exactly that much.

R=u2hgR = u\sqrt{\frac{2h}{g}}

Release distance from a moving platform

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)
WEBVTT

1
00:00:00.000 --> 00:00:05.000
Projectile from a moving platform — FemtoLearn.
2
00:00:05.000 --> 00:00:15.000
Released from a moving balloon/plane: the projectile inherits the platform's velocity.
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°