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Formulas/physics/Kinematics/Projectile from a Moving Body

Projectile from a Moving Body

Velocity of a projectile launched from a moving body — vector addition of body velocity and launch velocity.
Class 11Class JEE
Derivation

The situation

A body (a car, a plane, a train) is moving with velocity v⃗body\vec{v}_{body}. From this moving body, an object is launched with velocity v⃗rel\vec{v}_{rel} relative to the body. What is the actual velocity of the launched object relative to the ground?

v⃗projectile=v⃗body+v⃗rel\vec{v}_{projectile} = \vec{v}_{body} + \vec{v}_{rel}

The principle

This follows directly from the definition of relative velocity. If v⃗rel\vec{v}_{rel} is the velocity of the projectile as seen by someone on the moving body, and v⃗body\vec{v}_{body} is the velocity of the body relative to the ground, then the velocity of the projectile relative to the ground is the vector sum.

This is Galilean velocity addition — valid for speeds much less than the speed of light.

Examples

Horizontal throw from a moving car

A car moves at v0=20v_0 = 20 m/s east. A ball is thrown horizontally at 1010 m/s north relative to the car.

v⃗projectile=20i^+10j^ m/s\vec{v}_{projectile} = 20\hat{i} + 10\hat{j} \text{ m/s}

Speed relative to ground: 202+102=500≈22.4\sqrt{20^2 + 10^2} = \sqrt{500} \approx 22.4 m/s

The ball moves northeast relative to the ground, even though it was thrown north relative to the car.

Bomb dropped from a plane

A plane moves horizontally at 200200 m/s. A bomb is released (zero velocity relative to the plane at the moment of release).

v⃗bomb=200i^+0=200i^ m/s (horizontal)\vec{v}_{bomb} = 200\hat{i} + 0 = 200\hat{i} \text{ m/s (horizontal)}

The bomb has the same horizontal velocity as the plane at the moment of release. It then follows a parabolic path. From the ground, it moves forward while falling. From the plane, it appears to fall straight down.

Ball thrown upward from a moving train

A train moves at v0v_0 horizontally. A ball is thrown upward at speed vrelv_{rel} relative to the train.

Relative to the ground:

  • Horizontal velocity: v0v_0
  • Vertical velocity: vrelv_{rel} upward

The ball follows a parabolic trajectory relative to the ground. To a passenger on the train, the ball goes straight up and comes straight back down.

Trajectory from the ground vs from the moving body

ObserverWhat they see
On the moving bodyBall goes straight up (if thrown vertically relative to body)
On the groundBall follows a parabola

Both are correct — motion is always described relative to an observer. Neither is more "real" than the other.

Practical implication for problems

When a problem says "a projectile is launched from a moving vehicle", always:

  1. Find the initial velocity of the projectile relative to the ground using vector addition
  2. Use this velocity as uu (the launch speed) in the standard projectile equations
  3. Apply the standard kinematic analysis from that point
Key Idea
The key step is the vector addition at the moment of launch. After launch, the projectile is only under gravity (and air resistance if considered) — the motion of the original vehicle no longer affects the projectile in any way.
Note
A classic problem: a train moves at $v_0$ and a passenger throws a ball at speed $u$ at angle $\theta$ relative to the train. If the ball is to land back in the thrower's hand, it must go straight up relative to the train — meaning the launch angle relative to the ground is not $\theta$. The horizontal component of the ball's velocity relative to the ground must equal $v_0$.