Q.A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.

Concept: Relative Velocity – the path seen depends on the observer’s frame of reference.

1. In the car’s frame, the ball goes straight up and down because the boy and the ball share the car’s horizontal speed.
2. To the boy on the footpath (ground frame), the ball has the car’s constant horizontal velocity $v$ at the moment of toss, plus vertical motion due to gravity. …

Why Relative Velocity Matters

The key is understanding that motion looks different from different frames of reference. The boy in the car sees the ball go straight up and down because, relative to him, the ball has zero horizontal velocity at the moment of toss. But the boy on the footpath is stationary relative to the ground, so he sees the ball's actual motion: the ball already has the car's forward speed when released, and gravity acts only vertically.

This is a textbook example of Galilean relativity — the laws of motion are the same in all inertial frames, but the observed path depends on the observer's motion.

Step-by-Step Reasoning

1. Set up the frames of reference

Let the car move with constant speed $u$ along the positive $x$-direction. The boy in the car tosses the ball vertically upward with initial vertical speed $v_y$ (relative to himself). At the instant of toss, the ball's velocity relative to the ground is:

• Horizontal: $u$ (same as the car)
• Vertical: $v_y$ (upward)

2. Describe the forces after release

Once the ball leaves the boy's hand, the only force acting on it is gravity (downward). There is no horizontal force. So:

• Horizontal acceleration = $0$
• Vertical acceleration = $-g$ (taking upward as positive)

3. Write the equations of motion (ground frame)

Take the toss point as origin $(0,0)$ at time $t=0$.

• Horizontal position: $x(t) = u\,t$
• Vertical position: $y(t) = v_y t - \frac{1}{2} g t^2$

4. Eliminate time to get the path equation

From $x = u t$, we have $t = x/u$. Substitute into $y$:

This is of the form $y = ax - bx^2$, which is the equation of a parabola opening downward.

5. Interpret the shape

• The ball rises to a maximum height, then falls back. …

Concept: Transform a Known Trajectory Between Frames Instead of Re-Deriving It From Forces

Method: Galilean Shear of the Car-Frame Straight-Line Path

Rather than writing Newton's second law in the ground frame and eliminating time from scratch, this method starts from the trivial trajectory in the car's own frame (a straight vertical line) and applies the Galilean coordinate transformation directly to that curve -- turning a force-and-motion problem into a pure coordinate-geometry one.

Steps

1. Write the trivial trajectory in the car's frame. In the car's frame, the boy tosses the ball straight up with speed $v_y$ and no horizontal motion at all:

This is literally a vertical line segment -- no work needed here.

2. State the Galilean transformation between frames. The car frame moves at constant velocity $u$ relative to the ground, so a point's ground-frame coordinates relate to its car-frame coordinates by

3. Substitute the car-frame path into the transformation. Since $x'=0$ for every instant of the flight,

4. Eliminate $t$ using $t = x/u$ (identical algebra to before, but now applied to a shifted curve rather than derived from forces): $y = \frac{v_y}{u}x - \frac{g}{2u^2}x^2.$ …