A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s

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Q. A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.

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Concept understanding — Relative Motion

Relative Motion: Is It Really Moving?

Picture a parked car on a roundabout. To a pedestrian on the pavement, the car is perfectly still. But to a child riding the merry-go-round in the park across the street, the car appears to swing past in a wide circle as the ride spins. Who is right? Both are — because "moving" is never an absolute fact about an object. It only makes sense once you say moving relative to what.

The Idea: Every Motion Needs a Reference Frame

A reference frame is simply the object, point, or observer you choose to measure positions and motions against. When you say "the train is doing 80 km/h," you almost always mean relative to the ground — but that's a choice, not a law of nature. The ground itself is spinning with the Earth and racing around the Sun. There is no universal, built-in "zero" that everything else moves relative to.

Note

In classical mechanics this is sometimes called the relativity of motion: any two reference frames that move at a constant velocity relative to each other are equally valid for describing the laws of motion. Neither one is more "correct" than the other.

Once you accept this, a subtle but powerful consequence follows: position, velocity, and even the path an object appears to trace all depend on the observer's frame.

A Worked Illustration: The River and the Riverbank

Suppose a fish swims steadily across a flowing river. To someone standing on the riverbank, the fish's path looks like a diagonal line, curved by the current — it drifts downstream while crossing. But to a second observer floating alongside in a boat that moves with the current, the fish appears to swim in a perfectly straight line, directly across, with no drift at all.

Same fish, same instant, same water — two entirely different descriptions of the same motion, because the two observers are in different reference frames (one fixed to the bank, one moving with the water). Neither description is wrong.

Watch out

A common misconception is that one of these two descriptions must be "the truth" and the other an illusion. Both are equally real; they simply answer different questions — relative to the bank versus relative to the current.

Why This Isn't Just Philosophy

This idea isn't a curiosity — it's the starting point for solving real problems:

A pilot flying in a crosswind must think in two frames at once: the plane's motion relative to the air, and the air's motion relative to the ground, to know where the plane will actually end up.
An astronaut describes the Moon's motion relative to the Earth very differently from how someone describes it relative to the Sun — both descriptions are used, depending on the question being asked.
Two cars approaching each other on a highway each perceive the other as closing in much faster than either one's speedometer reads relative to the road.

Important

Before you can answer "how fast is it moving?" you must first answer "relative to what?" Skipping this question is the single most common source of confusion in relative-motion problems. …

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Since both balls experience the same acceleration due to gravity, their relative acceleration is zero. This means their relative velocity is constant and equal to their initial relative velocity. The relative speed of the balls is $\boxed{40 \text{ m/s}}$ .

When we talk about the speed or velocity of an object, it's always measured relative to some reference frame. In this problem, we are asked for the relative speed of the balls, which means the speed of one ball as observed from the frame of reference of the other ball.

The fundamental principle of relative motion states that the relative velocity of object A with respect to object B is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ . Similarly, the relative acceleration is $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$ .

The crucial insight here is that both balls are moving under the influence of gravity. The acceleration due to gravity, $g$ , acts downwards on both balls. Because both objects experience the same acceleration, their relative acceleration will be zero. If relative acceleration is zero, then the relative velocity must be constant.

Let's work through the steps:

Define a Coordinate System and Initial Conditions:

To analyze the motion, we first establish a consistent coordinate system. Let's choose the upward direction as positive. The acceleration due to gravity, $g$ , will therefore be negative.
- Ball 1 (dropped from building):
  
  Initial velocity, $u_1 = 0 \text{ m/s}$ (since it is dropped).
  
  Acceleration, $a_1 = -g$ (acting downwards).
- Ball 2 (thrown up from ground):
  
  Initial velocity, $u_2 = +40 \text{ m/s}$ (positive because it's thrown upwards).
  
  Acceleration, $a_2 = -g$ (acting downwards).
Calculate the Relative Acceleration:

The relative acceleration of ball 1 with respect to ball 2, denoted as $a_{12}$ , is the difference between their individual accelerations:

a_{12} = a_1 - a_2

Substituting the values for $a_1$ and $a_2$:

a_{12} = (-g) - (-g)

a_{12} = -g + g

a_{12} = 0

This result is fundamental: the acceleration of one ball relative to the other is zero.

> [!IMPORTANT]
> When two objects are moving solely under the influence of gravity (and neglecting air resistance), their relative acceleration is always zero. This is because gravity imparts the same acceleration ($g$) to all objects, regardless of their mass or initial velocity.

3. Determine the Relative Velocity: …

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Concept: The Freely-Falling Reference Frame Removes Gravity for Both Balls at Once

Method: The Equivalence-Principle Frame (Einstein's Falling Elevator) — See Both Balls as Moving at Constant Velocity, No Subtraction Needed

The stored answer writes $v_1(t)$ and $v_2(t)$ explicitly and algebraically subtracts them, watching the $-gt$ terms cancel. This method instead switches to a freely-falling reference frame — one that is itself accelerating downward at $g$ , as if it were a third object also in free fall (Einstein's "happiest thought": an observer in free fall feels no gravity at all) — in which BOTH balls appear to move at exactly their original launch velocities, forever, with no algebra required to see the relative velocity is constant.

Steps

Set up the falling frame. Let $S'$ be a reference frame released from rest at $t=0$ (the same instant both balls begin their motion) at the drop point, and which itself falls freely under gravity — its velocity relative to the ground is $v_{S'}(t) = -gt$ (taking upward as positive, since it's released from rest and falls under the same $g$ as everything else).
Find ball 1's velocity in this frame. Ball 1 is dropped from rest at $t=0$ , so its ground-frame velocity is $v_1(t) = -gt$ . In the falling frame:

v_1'(t) = v_1(t) - v_{S'}(t) = (-gt) - (-gt) = 0

Ball 1 appears permanently at rest in the falling frame — unsurprising, since it was released at the same instant, from the same point, under the same acceleration as the frame itself; it simply co-moves with $S'$ forever.

Find ball 2's velocity in this frame. Ball 2 is thrown upward at $t=0$ with speed $40\ \text{m/s}$ , so its ground-frame velocity is $v_2(t) = 40 - gt$ . In the falling frame:

v_2'(t) = v_2(t) - v_{S'}(t) = (40-gt) - (-gt) = 40 - gt + gt = 40\ \text{m/s}

The $-gt$ terms cancel identically, for every $t$ — not as an algebraic coincidence, but because both balls and the frame all share the exact same acceleration $-g$ . Ball 2 appears to move at a constant $40\ \text{m/s}$ upward in this frame, forever — this is exactly the equivalence principle in action: in a frame that is itself in free fall, gravity's effect on every other freely-falling object vanishes completely, and what remains is just each object's original launch velocity, unaccelerated.

Read off the relative velocity directly — no computation left to do. Since ball 1 is stationary ( $v_1'=0$ ) and ball 2 moves at a constant $40\ \text{m/s}$ in this frame, the relative velocity of ball 1 with respect to ball 2 (or vice versa) is simply:

v_{12}'(t) = v_1'(t) - v_2'(t) = 0 - 40 = -40\ \text{m/s}, \quad \text{constant for all } t

— the answer is visible immediately from Step 2 and 3 individually; no further subtraction, differentiation, or limit is needed. …

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Q.A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.

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