A ball is dropped from rest above the ground and bounces repeatedly. Its displacement $x$ (measured upward from the ground, all quantities taken positive upward) versus time $t$ is a succession of downward-opening parabolic arcs: starting at the maximum height at $t=0$, $x$ decreases to zero when the ball first strikes the ground, then rises to a second, lower peak, returns to zero, and so on — each successive arc reaching a smaller maximum height and lasting a shorter time. (a) Plot qualitatively the velocity versus time graph. (b) Plot qualitatively the acceleration versus time graph.

> [!TLDR] > During each flight the only force is gravity, so the acceleration is constant, $a=-g$, and the velocity changes linearly. At every bounce the ball reverses direction almost instantaneously, so $v$ jumps from a downward (negative) value to a smaller upward (positive) value. The $v$–$t$ graph is therefore a decaying saw-tooth of slope $-g$, and the $a$–$t$ graph is a constant $-g$ broken by brief upward spikes at each impact. ### Concept Taking up as positive, in free flight $a=-g$ (constant), so $$v(t)=v_{\text{start}}-g\,t,$$ a straight line of slope $-g$. The height is maximum where $v=0$ and the ball is at the ground ($x=0$) at each impact, where the direction of motion reverses. ### (a) Velocity–time graph - The ball is *dropped from rest*, so $v=0$ at $t=0$. - As it falls, $v$ becomes increasingly negative along a straight line of slope $-g$, reaching its most negative value $-v_1$ just before the first impact. - At the impact the velocity reverses almost instantly to a positive value $+v_2$ with $v_2<v_1$ (energy is lost), shown as a near-vertical upward jump. - It then decreases linearly (slope $-g$ again) through zero (top of the bounce) to $-v_2$ at the next impact, and so on. - The result is a saw-tooth: every sloping segment has the same slope $-g$, and the jump heights (bounce speeds) get smaller with each bounce. ### (b) Acceleration–time graph - Throughout every up-and-down flight, $a=-g$ (a horizontal line at $-g$). …

A ball is dropped from rest above the ground and bounces repeatedly. Its displacement x (measured upward from

NCERT Exemplar · Q17

x

(a) Plot qualitatively the velocity versus time graph.

(b) Plot qualitatively the acceleration versus time graph.

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Concept understanding — Motion Under Gravity

Motion Under Gravity – The Intuition First

Drop a ball. It falls. Why? Because the Earth pulls it. But the key insight is not just that it falls — it's how it falls. If you drop a stone from a cliff, does it move at a constant speed? No. It gets faster as it goes down. That's the heart of motion under gravity: the pull is steady, so the acceleration is steady.

Now imagine throwing a ball straight up. It rises, slows down, stops for an instant at the top, then falls back down. The same pull that made it slow down on the way up makes it speed up on the way down. The acceleration never changes direction — it always points downward, toward the centre of the Earth.

That constant downward acceleration is called $g$ . Near the Earth's surface, its magnitude is about $9.8\ \text{m/s}^2$ (often taken as $10\ \text{m/s}^2$ in exams for quick calculation). The direction is always downward.

Important

In one-dimensional motion under gravity, the acceleration is constant and equal to $g$ downward. This is true only when air resistance is negligible and the height is small compared to the Earth's radius.

The Precise Statement

Motion under gravity (in one dimension) means an object moves vertically under the sole influence of the Earth's gravitational pull. The acceleration is:

a = -g

where the negative sign indicates downward direction (if we take upward as positive). The value of $g$ is $9.8\ \text{m/s}^2$ (or $10\ \text{m/s}^2$ in many problems).

Because acceleration is constant, all three equations of uniformly accelerated motion apply directly:

v = u + at

s = ut + \frac12 at^2

v^2 = u^2 + 2as

But here $a$ is replaced by $-g$ (if upward is positive) or $+g$ (if downward is positive). The choice of sign convention is yours — just be consistent.

v = u - gt

Key Features You Must Know

1. Free Fall (dropped from rest)

If you simply let go of an object ( $u = 0$ ), it falls with increasing speed. After time $t$ , its velocity is $v = gt$ (downward). The distance fallen is $s = \frac12 gt^2$ .

2. Projected Upward

If you throw a ball upward with speed $u$ , it rises until its velocity becomes zero at the highest point. That takes time $t = u/g$ . The maximum height reached is $h = u^2/(2g)$ .

3. Symmetry of the Motion

The time to go up equals the time to come back down (to the same height). The speed at which it returns to the launch point equals the initial speed $u$ (but downward). This symmetry is a direct consequence of constant acceleration.

Watch out

At the highest point, velocity is zero but acceleration is still $g$ downward. Many beginners think acceleration becomes zero at the top — it does not. The object is momentarily at rest, but gravity is still pulling it.

A Simple Example …

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During each flight the only force is gravity, so the acceleration is constant, $a=-g$ , and the velocity changes linearly. At every bounce the ball reverses direction almost instantaneously, so $v$ jumps from a downward (negative) value to a smaller upward (positive) value. The $v$ – $t$ graph is therefore a decaying saw-tooth of slope $-g$ , and the $a$ – $t$ graph is a constant $-g$ broken by brief upward spikes at each impact.

Concept

Taking up as positive, in free flight $a=-g$ (constant), so

v(t)=v_{\text{start}}-g\,t,

a straight line of slope $-g$ . The height is maximum where $v=0$ and the ball is at the ground ( $x=0$ ) at each impact, where the direction of motion reverses.

(a) Velocity–time graph

The ball is dropped from rest, so $v=0$ at $t=0$ .
As it falls, $v$ becomes increasingly negative along a straight line of slope $-g$ , reaching its most negative value $-v_1$ just before the first impact.
At the impact the velocity reverses almost instantly to a positive value $+v_2$ with $v_2<v_1$ (energy is lost), shown as a near-vertical upward jump.
It then decreases linearly (slope $-g$ again) through zero (top of the bounce) to $-v_2$ at the next impact, and so on.
The result is a saw-tooth: every sloping segment has the same slope $-g$ , and the jump heights (bounce speeds) get smaller with each bounce.

(b) Acceleration–time graph

Throughout every up-and-down flight, $a=-g$ (a horizontal line at $-g$ ). …

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Concept: A Quantitative Geometric-Sequence Model of the Bouncing Ball (Coefficient of Restitution)

Method: Explicit Formulas via a Restitution Coefficient $e$ — a Quantitative Construction, Not a Qualitative Sketch

The stored answer describes the $v$ - $t$ and $a$ - $t$ graphs qualitatively ("a decaying saw-tooth," "brief spikes"). This method instead builds the graphs quantitatively, introducing a coefficient of restitution $e$ (the fraction of speed retained after each bounce, $0<e<1$ ) and writing an explicit formula for every impact speed, rebound speed, and flight duration — turning the qualitative description into a concrete, checkable sequence of numbers, the same style of construction used elsewhere in this chapter for repeated-bounce problems.

Setup

Let the ball drop from height $h$ , so its first impact speed is $v_1=\sqrt{2gh}$ (from rest). At each bounce, the ball's speed is reduced by the same factor $e$ (energy lost to the ground each time), so the rebound speed after the $k$ -th impact is $e^k v_1$ — and since each subsequent flight is a symmetric up-then-down trip (same speed up as the speed it comes back down with, gravity being the only force in flight), the speed just before the $(k+1)$ -th impact is also $e^k v_1$ .

Steps

Impact and rebound speed sequence (taking up as positive):

\text{impact speeds: } v_1,\ ev_1,\ e^2v_1,\ldots \qquad \text{rebound speeds: } ev_1,\ e^2v_1,\ e^3v_1,\ldots

Duration of the $k$ -th flight (the time between the $k$ -th and $(k+1)$ -th impact), from $2\times(\text{rebound speed})/g$ :

T_k = \frac{2\,e^k v_1}{g}, \qquad k=0,1,2,\ldots \quad (T_0 = v_1/g \text{ for the very first drop, half a "flight")}

Each successive flight is shorter than the last by exactly the factor $e$ — this is what makes the saw-tooth's segments shrink.

Velocity within each flight (slope always $-g$ , since only gravity acts):

v(t) = e^k v_1 - g\,(t - t_k), \qquad t_k \le t \le t_k+T_k

where $t_k$ is the start time of the $k$ -th flight. This confirms every sloping segment of the graph has the identical slope $-g$ , regardless of which bounce it follows — the quantitative version of the stored answer's "straight segments all of slope $-g$ ."

Jump height at each bounce — the discontinuity in $v$ at the $k$ -th impact:

\Delta v_k = (\text{rebound}) - (\text{impact}) = e^kv_1 - (-e^{k-1}v_1) = e^{k-1}v_1(1+e)

This jump shrinks geometrically by the factor $e$ from one bounce to the next — the quantitative reason the saw-tooth's amplitude visibly decays. …

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Concept

(a) Velocity–time graph

(b) Acceleration–time graph

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