A bob of mass $m$ hangs from a light string of length $L$ and is whirled in a vertical circle. Set up the circle in a vertical plane with centre O: point A is at the lowest point of the circle, point C is at the highest point (directly above A), point B is at the right-hand end of the horizontal diameter (at the same height as the centre), and point X is at the left-hand end of the horizontal diameter (also at the centre's height). If the string is cut, what will be the trajectory of the bob when it is cut at (a) point B, (b) point C, and (c) point X?

A bob of mass m hangs from a light string of length L and is whirled in a vertical circle. Set up the circle i

NCERT Exemplar · Q33

m

(a) point B,

(b) point C, and

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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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When the string is cut the bob flies off along the tangent to the circle at that point and then moves as a projectile under gravity. At B and X the tangent is vertical, so the bob travels along a vertical straight line; at the top C the tangent is horizontal, so the bob follows a parabolic path.

Key idea

The velocity of the bob is always tangent to the circle. After the string is cut, the only force is gravity (vertical). The shape of the path depends on the direction of this tangential velocity.

(a) String cut at B (right end of horizontal diameter)

The radius OB is horizontal, so the tangent — and hence the velocity — is vertical. Velocity and gravity are both vertical (same line), so there is no horizontal motion: the bob moves along a vertical straight line (it rises, slows and falls back along that line, or moves straight down, depending on the sense of rotation).

(b) String cut at C (top) …

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Key idea

(a) String cut at B (right end of horizontal diameter)

(b) String cut at C (top) …

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