A football is kicked into the air vertically upwards. What is its (a) acceleration, and (b) velocity at the hi

NCERT Exemplar · Q19

Q. A football is kicked into the air vertically upwards. What is its

(a) acceleration, and

(b) velocity at the highest point?

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Concept understanding — Projectile Motion Under Gravity

Projectile Motion Under Gravity

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it curves upward, then arcs downward. That curve is a parabola, and the motion is called projectile motion.

The key insight: once the ball leaves your hand, the only force acting on it (ignoring air resistance) is gravity pulling it straight down. There is no force pushing it sideways or upward after release. That single downward force is what creates the beautiful curved path.

The Core Idea

A projectile is any object that is thrown, launched, or otherwise projected into the air and then moves under the influence of gravity alone. The motion has two independent parts happening simultaneously:

Horizontal motion: constant speed (no horizontal force)
Vertical motion: constant downward acceleration $g \approx 9.8 \, \text{m/s}^2$

These two motions are completely independent — they don't affect each other. This is the most important thing to understand.

Important

The horizontal and vertical motions are independent. The horizontal speed stays constant; the vertical speed changes by $9.8 \, \text{m/s}$ every second downward.

Breaking It Down Mathematically

Let's set up coordinates: $x$ is horizontal, $y$ is vertical (positive upward). The launch point is at $(0,0)$ with initial speed $u$ at angle $\theta$ above horizontal.

Initial velocity components:

u_x = u \cos \theta

u_y = u \sin \theta

Horizontal motion (no acceleration):

x = u_x t = (u \cos \theta) t

Vertical motion (constant downward acceleration $g$ ):

y = u_y t - \frac{1}{2} g t^2 = (u \sin \theta) t - \frac{1}{2} g t^2

The minus sign is because gravity pulls downward, opposite to our positive $y$ direction.

The Path Is a Parabola

Eliminate $t$ between the $x$ and $y$ equations. From $x = u_x t$ , we get $t = \frac{x}{u \cos \theta}$ . Substitute into the $y$ equation:

y = (u \sin \theta) \left( \frac{x}{u \cos \theta} \right) - \frac{1}{2} g \left( \frac{x}{u \cos \theta} \right)^2

y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2

This is of the form $y = ax - bx^2$ , which is a parabola opening downward. That's why every projectile under gravity follows a parabolic path.

y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2

Key Quantities You'll Need

Time of Flight ( $T$ )

The total time the projectile stays in the air. Set $y = 0$ (returns to launch height):

0 = (u \sin \theta) T - \frac{1}{2} g T^2

Factor $T$ : $T (u \sin \theta - \frac{1}{2} g T) = 0$

The non-zero solution:

T = \frac{2 u \sin \theta}{g}

Maximum Height ( $H$ )

The highest point occurs when vertical velocity becomes zero: $v_y = u \sin \theta - g t = 0$ , so $t = \frac{u \sin \theta}{g}$ .

Plug into $y$ equation:

H = (u \sin \theta) \left( \frac{u \sin \theta}{g} \right) - \frac{1}{2} g \left( \frac{u \sin \theta}{g} \right)^2

H = \frac{u^2 \sin^2 \theta}{g} - \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin^2 \theta}{2g}

Range ( $R$ )

Horizontal distance traveled when it returns to launch height. Use $x = u_x T$ :

R = (u \cos \theta) \cdot \frac{2 u \sin \theta}{g} = \frac{2 u^2 \sin \theta \cos \theta}{g}

Using $\sin 2\theta = 2 \sin \theta \cos \theta$ :

R = \frac{u^2 \sin 2\theta}{g}

Tip

Maximum range occurs when $\sin 2\theta = 1$ , i.e., $2\theta = 90^\circ$ or $\theta = 45^\circ$ . At this angle, $R_{\text{max}} = \frac{u^2}{g}$ .

Common Mistakes to Avoid

Watch out

Don't mix up horizontal and vertical equations. Horizontal has constant speed; vertical has constant acceleration. …

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At every point of its flight—including the highest point—the ball experiences constant downward acceleration $g \approx 9.8\,\text{m/s}^2$ due to gravity. The velocity at the highest point is zero (the ball stops momentarily before falling back), but the acceleration never vanishes.

Why acceleration persists even when velocity is zero

A common intuition says "if something stops, nothing is acting on it." That intuition fails here. Acceleration measures how quickly velocity changes, not whether the object is moving. Gravity pulls on the football throughout its journey—on the way up, at the peak, and on the way down—so the acceleration is the same everywhere: $\vec{a} = -g\,\hat{j}$ (taking upward as positive).

The highest point is special only because the velocity passes through zero there. The ball decelerates (slows down) while rising, reaches zero speed at the top, then accelerates downward. The agent causing that deceleration and subsequent fall is gravity, which never switches off.

Step-by-step reasoning

Identify the force acting on the ball. Once the ball leaves the foot, the only force is gravity (we neglect air resistance). By Newton's second law, $\vec{F} = m\vec{a}$ , so

\vec{a} = \frac{\vec{F}_{\text{gravity}}}{m} = \frac{-mg\,\hat{j}}{m} = -g\,\hat{j}.

This acceleration is constant in magnitude and direction throughout the flight.

Recognize that acceleration does not depend on velocity.

Gravity acts whether the ball is moving fast, slow, or not at all. At the highest point the velocity is zero, but $\vec{F} = m\vec{a}$ still holds with $\vec{F} = -mg\,\hat{j}$ , so $\vec{a} = -g\,\hat{j}$ there too.
Find the velocity at the highest point.

"Highest point" means the ball has stopped rising and is about to fall. At that instant the vertical component of velocity is zero:

$v_{\text{top}} = 0.$ …

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Concept: Read Both Answers Off a Single $v$ - $t$ Line

Method: Velocity-Time Graph (Slope = Acceleration; Zero-Crossing = the Highest Point) -- No Force Argument Needed

Rather than invoking Newton's second law and separately arguing "acceleration doesn't switch off," this method sketches the vertical velocity $v(t)=u-gt$ as a straight line on a $v$ - $t$ graph and reads both answers directly off its geometry: the graph's constant slope gives the acceleration everywhere, and the graph's zero-crossing gives the velocity at the highest point, as a single picture.

Step 1 -- Plot $v(t) = u - gt$ as a straight line

Taking upward as positive, with $u$ the launch speed: at $t=0$ , $v=u>0$ (moving up); as $t$ increases, $v$ decreases linearly, crosses zero at some time $t^\ast$ , and continues into negative values (moving down) beyond that. This is one continuous straight line on the $v$ - $t$ plane, with no kink, break, or change in slope anywhere -- including right at the point where it crosses the time axis.

Step 2 -- Read acceleration off the slope, everywhere on the line

The slope of a $v$ - $t$ graph is, by definition, the acceleration:

\text{slope} = \frac{dv}{dt} = -g

Since the graph is one unbroken straight line with a single constant slope, the acceleration equals $-g$ (magnitude $g$ , downward) at literally every point on the graph -- including the point where $v=0$ . A straight line's slope does not change value just because the line happens to cross the horizontal axis at that point; the slope is a property of the whole line, not of any one point on it.

Step 3 -- Read the velocity at the highest point directly as the graph's zero-crossing

"Highest point" is, by definition, the instant the ball stops rising and is about to start falling -- exactly the instant the $v$ - $t$ graph crosses from positive to negative, i.e. its zero-crossing. Solving $0 = u-gt^\ast$ gives $t^\ast=u/g$ , and by construction of the graph, $v(t^\ast)=0$ . …

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Q.A football is kicked into the air vertically upwards. What is its

Why acceleration persists even when velocity is zero

Step-by-step reasoning

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