A ball is thrown vertically upward with an initial velocity of 20 m/s. Taking $g = 10\ \text{m/s}^2$, find (a) the maximum height reached and (b) the total time taken by the ball to return to the point of projection.

**Setting up.** Take the upward direction as positive. Initial velocity $u = 20$ m/s (upward), and since gravity acts downward, the acceleration is $a = -g = -10\ \text{m/s}^2$ throughout the flight. **(a) Maximum height.** At the highest point, the ball's velocity is momentarily zero ($v=0$, see Exercise 2). Using $v^2 = u^2 + 2as$: $$0 = (20)^2 + 2(-10)h \quad\Longrightarrow\quad 0 = 400 - 20h \quad\Longrightarrow\quad h = \frac{400}{20} = 20\ \text{m}$$ **(b) Time to return to the point of projection.** First find the time to reach the highest point, using $v = u + at$ with $v=0$: $$0 = 20 + (-10)t_{\text{up}} \quad\Longrightarrow\quad t_{\text{up}} = \frac{20}{10} = 2\ \text{s}$$ By the symmetry of motion under constant gravity (ignoring air resistance), the time taken to fall back down from the highest point to the starting height equals the time taken to rise, so $t_{\text{down}} = t_{\text{up}} = 2$ s. The total time of flight is therefore $$T = t_{\text{up}} + t_{\text{down}} = 2 + 2 = 4\ \text{s}$$ **Cross-check.** Using $s = ut + \tfrac{1}{2}at^2$ with $s=0$ (return to the same height) and $u=20$, $a=-10$: $0 = 20T - 5T^2 = 5T(4-T)$, giving $T=0$ (the start) or $T=4$ s (the return) — confirming the answer. [!ANSWER] Maximum height reached = 20 m; total time to return to the point of projection = 4 s.

A ball is thrown vertically upward with an initial velocity of 20 m/s. Taking g = 10m/s², find (a) the maximum

Numerical · Q15

g = 10\ \text{m/s}^2

(a) the maximum height reached and

(b) the total time taken by the ball to return to the point of projection.

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Concept understanding — Uniformly Accelerated Motion

Uniformly Accelerated Motion

Imagine you're sitting in a train that starts moving from a station. At first, it crawls — then it picks up speed smoothly, second by second. If you watch the speedometer, you might see it climb by the same amount every second: 0 to 10 km/h, then 10 to 20, then 20 to 30. That steady, predictable increase is the heart of uniformly accelerated motion.

The Intuition

When something moves with uniform acceleration, its velocity changes by the same amount in every equal interval of time. The change is constant — not faster one second and slower the next.

Think of a ball rolling down a gentle, straight ramp. It starts from rest. In the first second, it gains some speed. In the next second, it gains exactly the same amount of speed again. The acceleration — the rate of change of velocity — is fixed.

Note

"Uniform" here means "constant" or "unchanging." It does not mean the speed is constant. In fact, the speed is changing — but the rate at which it changes is constant.

The Precise Statement

Uniformly accelerated motion is motion in a straight line where the acceleration $a$ is constant in both magnitude and direction.

Mathematically, if $v$ is velocity at time $t$ , and $u$ is the initial velocity (at $t = 0$ ), then:

a = \frac{v - u}{t} = \text{constant}

This single idea leads to the three famous equations of motion (for constant acceleration):

v = u + at

Here:

$u$ = initial velocity (at $t = 0$ )
$v$ = velocity at time $t$
$a$ = constant acceleration
$s$ = displacement in time $t$

What It Looks Like in Real Life

Situation	Acceleration	Why it's (approximately) uniform
A car accelerating on a highway	~2–3 m/s²	Engine provides roughly constant force
A ball dropped from a height	9.8 m/s² downward	Gravity is nearly constant near Earth's surface
A train starting from a station	~0.5 m/s²	Controlled by the driver to be smooth

Watch out

Not all motion is uniformly accelerated. A car stopping suddenly has deceleration that changes — it's not uniform. A roller coaster has acceleration that varies wildly. Uniform acceleration is an ideal model that works beautifully for many real situations (like free fall) but not all.

The Key Insight

The word "uniform" refers to the acceleration, not the velocity. If acceleration is constant, then:

Velocity changes linearly with time (a straight line on a $v$ - $t$ graph)
Displacement changes quadratically with time (a parabola on an $s$ - $t$ graph)

This is why the equations above are so powerful: they let you predict position and velocity at any instant, as long as acceleration stays constant.

Important

For uniformly accelerated motion, the $v$ - $t$ graph is always a straight line. The slope of that line equals the acceleration. If the graph is curved, acceleration is not uniform.

A Simple Example to Cement It

A car starts from rest ( $u = 0$ ) and accelerates uniformly at $2 \text{ m/s}^2$ for 5 seconds.

After 1 s: $v = 0 + 2(1) = 2 \text{ m/s}$
After 2 s: $v = 0 + 2(2) = 4 \text{ m/s}$
After 3 s: $v = 6 \text{ m/s}$
After 4 s: $v = 8 \text{ m/s}$
After 5 s: $v = 10 \text{ m/s}$

Every second, the speed increases by exactly 2 m/s. That's uniform acceleration.

The distance covered in those 5 seconds? Using $s = ut + \frac{1}{2}at^2$ :

s = 0 + \frac{1}{2}(2)(5^2) = \frac{1}{2} \times 2 \times 25 = 25 \text{ meters}

So the car travels 25 m while smoothly picking up speed to 10 m/s.

The Bottom Line

Uniformly accelerated motion = constant acceleration. Velocity changes by equal amounts in equal times. It's the simplest kind of accelerated motion, and it's the foundation for understanding everything from falling apples to rocket launches.

Uniformly Accelerated Motion is a central idea in the NCERT Class 11 Physics chapter on Motion in a Straight Line, matching searches like "uniformly accelerated motion equations" or "kinematics important questions class 11 physics". The three equations of motion derived here are tested constantly across CBSE boards and are an important topic for JEE Main and NEET, building on this NCERT Class 11 Physics foundation.

Setting up. Take the upward direction as positive. Initial velocity $u = 20$ m/s (upward), and since gravity acts downward, the acceleration is $a = -g = -10\ \text{m/s}^2$ throughout the flight.

Maximum height. At the highest point, the ball's velocity is momentarily zero ( $v=0$ , see Exercise 2). Using $v^2 = u^2 + 2as$ : $0 = (20)^2 + 2(-10)h \quad\Longrightarrow\quad 0 = 400 - 20h \quad\Longrightarrow\quad h = \frac{400}{20} = 20\ \text{m}$
Time to return to the point of projection. First find the time to reach the highest point, using $v = u + at$ with $v=0$ : $0 = 20 + (-10)t_{\text{up}} \quad\Longrightarrow\quad t_{\text{up}} = \frac{20}{10} = 2\ \text{s}$ By the symmetry of motion under constant gravity (ignoring air resistance), the time taken to fall back down from the highest point to the starting height equals the time taken to rise, so $t_{\text{down}} = t_{\text{up}} = 2$ s. The total time of flight is therefore $T = t_{\text{up}} + t_{\text{down}} = 2 + 2 = 4\ \text{s}$ Cross-check. Using $s = ut + \tfrac{1}{2}at^2$ with $s=0$ (return to the same height) and $u=20$ , $a=-10$ : $0 = 20T - 5T^2 = 5T(4-T)$ , giving $T=0$ (the start) or $T=4$ s (the return) — confirming the answer. [!ANSWER] Maximum height reached = 20 m; total time to return to the point of projection = 4 s.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set SET-AP55001 mark
$Q. The rate of change of velocity of an object is called ________.$
›Reveal solutionSolution

The rate of change of velocity of an object with respect to time is called its acceleration.

Velocity can change in magnitude (speeding up/slowing down), direction, or both. Acceleration is defined as:

a = dv/dt = (change in velocity)/(time taken)

It is a vector quantity, with SI unit m/s^2. If velocity increases, acceleration is in the direction of motion; if velocity decreases, acceleration is opposite to the direction of motion (often called retardation/deceleration in that case).

✓Final answer
$Acceleration.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Velocity of an object moving with uniform acceleration will be:$
(a) decreasing
(b) increasing
(c) will be zero
(d) increasing or decreasing
›Reveal solutionSolution

With uniform acceleration the velocity may either increase or decrease.

Uniform acceleration means the velocity changes by equal amounts in equal times. If the acceleration is in the same direction as the velocity, the speed increases; if it is opposite (retardation), the speed decreases.

Hence the velocity of a uniformly accelerated body can be increasing or decreasing.

✓Final answer
$(D) increasing or decreasing.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. An object is moving with a constant acceleration. Its velocity time graph is:$
(a) graph A
(b) graph B
(c) graph C
(d) graph D
›Reveal solutionSolution
Constant acceleration means a straight-line v–t graph with constant slope — that's graph (A).

Reasoning. For constant acceleration $a$ , the equation of motion is $v = u + at$ , which is linear in $t$ (a straight line with slope $a$ and intercept $u$ ).

Checking each sketch:

(A) Straight line from the origin with constant positive slope — velocity increases uniformly with time, exactly $v=at$ (starting from rest). This matches constant acceleration.

(B) Horizontal line — velocity doesn't change with time, so acceleration is zero, not constant nonzero acceleration.

(C) A vertical line — implies velocity changing instantaneously with no time elapsed, i.e. infinite acceleration, which is unphysical.

(D) An upward-curving (exponential-like) curve — the slope keeps increasing, meaning the acceleration itself is increasing with time, i.e. NOT constant.

Only a straight line with fixed nonzero slope, as in (A), represents uniformly (constantly) accelerated motion.

✓Final answer
$Graph (A) is the correct v-t graph for constant acceleration.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. If the slope of the velocity time graph of a particle moving in stright line be constant, then its motion is with —$
(a) Uniform velocity
(b) Uniform acceleration
(c) Variable acceleration
(d) Constant speed
›Reveal solutionSolution

Constant slope of a v–t graph ⇒ uniform acceleration.

Acceleration $a = \dfrac{dv}{dt}$ = slope of the velocity–time graph.

If this slope is constant, then a is constant in both magnitude and direction, i.e. the body moves with uniform acceleration (a straight-line v–t graph).

✓Final answer
$(b) Uniform acceleration.$
CBSE 2021Set ANNUAL1 markMCQ
$Q. The distance travelled by a particle is directly proportional to the square of the time taken, the acceleration of the particle is :$
(a) Increasing
(b) Decreasing
(c) Zero
(d) Constant
›Reveal solutionSolution

$s = kt^2$ differentiates to a constant acceleration $a = 2k$ .

Given $s \propto t^2$ , write $s = k t^2$ for some constant $k$ .

Velocity: $v = \dfrac{ds}{dt} = 2kt$

Acceleration: $a = \dfrac{dv}{dt} = 2k$

Since $k$ is a constant, $a = 2k$ does not depend on time — it is constant. (This is exactly the case of uniformly accelerated motion from rest, where $s = \tfrac{1}{2}at^2$ .)

✓Final answer
$The acceleration is constant \to option (d).$
CBSE 2019Set ANNUAL1 markMCQ
$Q. Which graph represents uniform acceleration ?$
(a) graph
(a) -- straight line through origin, constant slope
(b) graph
(b) -- triangular rise-then-fall
(c) graph
(c) -- concave-up increasingly steep curve
(d) graph
(d) -- concave-down flattening/saturating curve
›Reveal solutionSolution
A displacement-time graph for uniform acceleration is a concave-up curve that gets steeper with time, since s depends on t^2 -- graph (c).

For motion with uniform (constant) acceleration a, the displacement-time relation is

s = ut + (1/2)a t^2

This is a quadratic function of time. Its slope, ds/dt = u + at, is the instantaneous velocity, which increases steadily with time for positive a. A steadily increasing slope means the curve bends upward more and more sharply as t increases -- this is exactly a concave-up curve that starts at the origin and curves increasingly steeply, matching graph (c).

By contrast:

a straight line through the origin with constant slope describes UNIFORM VELOCITY (zero acceleration), not uniform acceleration.

a rise-then-fall triangular shape describes a velocity-time graph of a particle that accelerates then decelerates, not a displacement-time graph of uniform acceleration.

(d) a curve that flattens out (concave down, saturating) describes motion where velocity decreases with time, i.e., a deceleration or approach to a terminal/equilibrium value, not uniform acceleration.

✓Final answer
$The correct option is (c) graph (c) -- concave-up increasingly steep curve.$

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Q.A ball is thrown vertically upward with an initial velocity of 20 m/s. Taking g=10 m/s2g = 10\ \text{m/s}^2g=10 m/s2, find

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Q.A ball is thrown vertically upward with an initial velocity of 20 m/s. Taking $g = 10\ \text{m/s}^2$ , find

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