Q.A boy throws a ball in air at $60^\circ$ to the horizontal along a road with a speed of 10 m/s (36km/h). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of (18km/h). Give explanation to support your diagram.

Concept: Relative Velocity

The boy in the car observes the ball's motion in the car's reference frame. We must subtract the car's velocity from the ball's velocity to find the ball's velocity relative to the car.

Step 1: Convert speeds and resolve the ball's velocity.

Car speed: $v_c = 18 \text{ km/h} = 5 \text{ m/s}$ (horizontal, along the road).

Ball's velocity components: $v_{bx} = 10\cos 60° = 5 \text{ m/s}$ (horizontal), $v_{by} = 10\sin 60° = 5\sqrt{3} \text{ m/s}$ (vertical).

Step 2: Find relative velocity.

Horizontal component relative to car: $v_{rel,x} = v_{bx} - v_c = 5 - 5 = 0 \text{ m/s}$.

Vertical component: $v_{rel,y} = v_{by} = 5\sqrt{3} \text{ m/s}$ (unchanged).

Step 3: Interpret the motion. …

Understanding Relative Velocity

When two observers move relative to each other, they see the same object trace different paths. The boy in the car sees the ball's motion relative to his own frame of reference. To find this, we subtract the car's velocity vector from the ball's velocity vector.

The key insight: if the horizontal component of the ball's velocity exactly matches the car's speed, the ball will appear to have no horizontal motion relative to the car. It will seem to go straight up and straight down.

Step-by-Step Analysis

1. Resolve the ball's initial velocity into components.

   The ball is thrown at $60^\circ$ to the horizontal with speed $v_0 = 10 \text{ m/s}$.

2. Identify the car's velocity.

   The car moves horizontally along the road at $18 \text{ km/h} = 5 \text{ m/s}$.

3. Calculate the ball's velocity relative to the car.

   The relative velocity is:

Horizontal component:

Vertical component:

4. Interpret the relative motion.

   The ball has zero horizontal velocity relative to the car. From the car's frame, the ball appears to move only vertically. It rises with initial upward velocity $5\sqrt{3} \text{ m/s}$, decelerates under gravity, reaches a maximum height, then falls back down along the same vertical line.

The Sketch …

Concept: Transform the Ground-Frame Trajectory Equation Itself, Not Just the Velocity Components

Method: Substitute a Moving Coordinate Into the Standard Projectile Equation

The direct method subtracts velocity vectors at the instant of launch and reasons about the resulting constant relative velocity. This method instead writes the ball's full trajectory equation in the ground frame first, then substitutes the car's moving coordinate directly into it -- confirming the same conclusion from the shape of the path rather than from an instantaneous velocity snapshot.

Steps

1. Write the ball's standard trajectory in the ground frame. Launched at speed $v_0=10\text{ m/s}$, angle $60^\circ$:

2. Write the car's position as a function of time. The car moves at $18\text{ km/h}=5\text{ m/s}$, so $x_{\text{car}}(t)=5t$.

3. Define the car-frame horizontal coordinate $x' = x - x_{\text{car}} = x - 5t$. Using the ball's own horizontal motion $x = v_0\cos60^\circ\,t = 5t$ (since $v_0\cos60^\circ=5\text{ m/s}$):

4. Recognize the coincidence is structural, not a special value. Because the ball's horizontal ground-speed ($5\text{ m/s}$) is built from the same $\cos60^\circ$ factor that produces that number, and the car's speed was chosen to match it exactly, $x'(t)\equiv 0$ identically -- so no matter what the vertical motion does, the ball's car-frame $x$-coordinate never changes. …