In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajecto

NCERT Exemplar · Q23

Q. In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

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Concept understanding — Projectile Motion

Projectile Motion — From Intuition to Precision

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it rises, slows down, then curves downward and falls. That curved path is a projectile's trajectory. The ball is a projectile: any object that is launched into the air and then moves only under the influence of gravity (and air resistance, which we ignore for now).

The key intuition: once the ball leaves your hand, the only force acting on it is gravity pulling it straight down. There is no forward force after release. The ball keeps moving forward because of inertia — it wants to keep going in a straight line at constant speed. But gravity keeps pulling it down, so the forward motion and downward acceleration combine to produce a curved path.

The Precise Statement

g \approx 9.8 \, \text{m/s}^2

We break the motion into two independent components:

Horizontal motion: No acceleration ( $a_x = 0$ ). So horizontal velocity $v_x$ is constant.
Vertical motion: Constant downward acceleration ( $a_y = -g$ ). So vertical velocity $v_y$ changes linearly with time.

The independence of these components is the central idea — what happens vertically does not affect what happens horizontally, and vice versa.

The Equations (for a projectile launched with initial speed $u$ at angle $\theta$ above horizontal)

First, resolve the initial velocity:

u_x = u \cos\theta, \quad u_y = u \sin\theta

Horizontal motion (constant velocity):

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

Vertical motion (constant acceleration $-g$ ):

v_y = u_y - gt = u \sin\theta - gt

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

Key Results You Must Know

Important

Time of flight $T$ : total time the projectile stays in the air (until $y=0$ again).

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

Important

Maximum height $H$ : the highest vertical position reached (when $v_y = 0$ ).

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

Important

Range $R$ : the horizontal distance covered when it returns to launch height.

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

Note

The range is maximum when $\sin 2\theta = 1$ , i.e., $\theta = 45^\circ$ . For a given speed, $45^\circ$ gives the farthest throw.

The Trajectory Equation (Path Shape)

Eliminate $t$ from the $x$ and $y$ equations to get $y$ as a function of $x$ :

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

This is a parabola — the signature shape of projectile motion.

Common Mistake to Avoid …

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Air resistance opposes motion at every point, so the projectile loses speed continuously and its path becomes asymmetric — the descent is steeper and shorter than the ascent, and the range is reduced. The trajectory is not a parabola.

Why the parabola is an idealisation

When we ignore air resistance, the only force is gravity — constant, vertical, downward. That gives a horizontal velocity that never changes and a vertical acceleration of $g$ . The maths then yields $y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}$ , which is a perfect parabola. Symmetric ascent and descent, maximum range at $45^\circ$ , all that.

But in the real world, air resistance (drag) is always present. It acts opposite to the instantaneous velocity vector, so it has both horizontal and vertical components. That changes everything.

How drag reshapes the motion

Drag force depends on speed. For typical projectiles (not too small, not too fast), drag is roughly proportional to $v^2$ : $F_d = \frac12 C_d \rho A v^2$ . The key point: it always points opposite to $\vec{v}$ .
Horizontal motion is no longer uniform. Without drag, $v_x$ stays constant. With drag, there is a horizontal component of $F_d$ that continuously slows the projectile. So the horizontal speed keeps decreasing, and the projectile takes less time to cover the same horizontal distance on the way down than it did on the way up.
Vertical motion is altered on both legs. On the way up, drag adds to gravity — both act downward — so the projectile decelerates faster than $g$ alone. On the way down, drag opposes gravity (points upward), so the net downward acceleration is less than $g$ . But because the projectile has lost so much speed during ascent, the descent is still quicker overall.
The asymmetry. On the way up, the projectile has high speed, so drag is large. It loses energy rapidly. By the time it reaches the top, its speed is much lower than the initial speed. On the way down, it starts from that low speed, and drag (now upward) is small. So the descent is steeper and shorter in horizontal extent than the ascent.

Watch out

A common mistake is to think drag only affects the range. It also destroys the symmetry of the path — the apex is not at the midpoint of the horizontal range, and the angle of impact is steeper than the launch angle.

Sketching the trajectory

Draw a standard parabolic arc for reference (dashed). Then draw the real path: …

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Concept: Solve an Exactly-Solvable Simplified Drag Model Instead of Reasoning Qualitatively

Method: Closed-Form Solution for Linear (Stokes) Drag

Real air resistance on a macroscopic projectile is usually closer to $F_d\propto v^2$ , which has no elementary closed-form trajectory. But the linear-drag case, $F_d = -k\vec v$ (Stokes drag), is exactly solvable and reproduces the same qualitative asymmetry -- so solving it explicitly gives a rigorous, checkable picture of the shape instead of relying on force-direction intuition alone.

Steps

Write the equations of motion with linear drag, defining the time constant $\tau = m/k$ :

m\dot v_x = -kv_x \implies \dot v_x = -\frac{v_x}{\tau}, \qquad m\dot v_y = -mg-kv_y \implies \dot v_y = -g - \frac{v_y}{\tau}.

Solve the horizontal equation (simple exponential decay):

v_x(t) = v_{0x}\,e^{-t/\tau}.

Solve the vertical equation (linear ODE with constant forcing $-g$ ); the steady-state (terminal) solution is $v_y\to -g\tau$ , so:

v_y(t) = (v_{0y}+g\tau)\,e^{-t/\tau} - g\tau.

Integrate to get positions:

x(t) = v_{0x}\tau\left(1-e^{-t/\tau}\right), \qquad y(t) = -g\tau t + (v_{0y}+g\tau)\tau\left(1-e^{-t/\tau}\right).

Read off the shape. As $t\to\infty$ , $x(t)\to v_{0x}\tau$ -- the horizontal range approaches a finite limiting value instead of growing forever, so the trajectory bends toward a near-vertical asymptote. Meanwhile $v_y(t)\to -g\tau$ (terminal velocity), so the descent eventually becomes a constant-speed near-vertical fall -- exactly the "falls almost straight down" behaviour sketched qualitatively. …

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Q.In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

Why the parabola is an idealisation

How drag reshapes the motion

Sketching the trajectory

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