An object thrown from a moving bus is an example of
(A) Uniform circular motion
(B) Rectilinear motion
(C) Projectile motion
(D) Motion in one dimension

Question

Accepted Answer

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

Watch out

Many students think the projectile has zero velocity at the highest point. Wrong. At the top, $v_y = 0$ , but $v_x$ is still $u \cos\theta$ — the projectile is still moving horizontally. It never stops; it just stops rising.

Why This Matters

Projectile motion is the foundation for understanding any motion under constant force in two dimensions — from a cricket ball to a rocket stage separation (though rockets have thrust). In exams, you will often be asked to:

Find time of flight, height, or range given launch conditions.
Work backwards from a given range or height to find launch speed or angle.
Solve problems where launch and landing are at different heights (e.g., from a cliff).

Master the independence of horizontal and vertical motion, and you have the key to every projectile problem.

For quick revision, remember that Projectile Motion is a core, NCERT-aligned topic from the Motion in a Plane (Projectile Motion) portion of the Class 11 Physics curriculum, and it is tested regularly in CBSE board exams as well as in JEE Main and NEET, which is exactly why "Projectile Motion important questions" shows up so often in Physics question banks. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.

The bus itself may be moving in a straight line, but the question is about the motion of the object after it leaves the bus. From that instant, the only force on it is gravity (air resistance neglected), which is exactly the situation section 3.3.5 defines as projectile motion — an object launched with some initial velocity (here, the velocity it had while still on the moving bus) that then moves only under Earth's gravitational field, tracing a parabolic path. [!ANSWER] (C) Projectile motion

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

CBSE 2026Set ANNUAL1 markMCQ
$Q. The value of acceleration at the highest point of the projectile path is$
(a) zero
(b) maximum
(c) minimum
(d) equal to g
›Reveal solutionSolution

Throughout projectile motion the acceleration equals g, so at the top it is also g. Answer (D).

In projectile motion (ignoring air resistance) the only force is gravity. Hence the acceleration is constant and equals g (about 9.8 m/s^2), pointing vertically downward, at all instants.

At the highest point the VERTICAL velocity becomes zero, but the acceleration does not; it is still g. (Only the velocity is horizontal there, not the acceleration.)

✓Final answer
$(D) equal to g.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The maximum limit (range) of a gun is 3920 m. What is the velocity of shoot? (A) 200 m/s (B) 196 m/s (C) 100 m/s (D) 150 m/s$
›Reveal solutionSolution

The gun's muzzle speed is 196 m/s, found from the maximum-range formula of projectile motion.

The range of a projectile launched with speed $u$ at angle $\theta$ is $R = \dfrac{u^2\sin2\theta}{g}$ . The range is maximum when $\theta = 45^\circ$ (so $\sin2\theta = 1$ ), giving:

$R_{max} = \dfrac{u^2}{g}$

Given $R_{max} = 3920$ m and $g = 9.8\text{ m/s}^2$ :

$u^2 = R_{max}\times g = 3920 \times 9.8 = 38416$

$u = \sqrt{38416} = 196\text{ m/s}$

✓Final answer
$(B) 196 m/s.$
CBSE 2024Set SET-AP55001 mark
$Q. The path of projectile motion is ________.$
›Reveal solutionSolution

The trajectory traced by a projectile (launched at an angle to the horizontal, under gravity alone) is a parabola.

For a projectile launched with initial speed u at angle θ, the horizontal motion has constant velocity u cos θ (no horizontal force, ignoring air resistance), so x = (u cos θ) t. The vertical motion is uniformly accelerated by gravity, so y = (u sin θ) t − ½gt^2.

Eliminating t (t = x / (u cos θ)) and substituting into the y-equation gives:

y = x tan θ − [g / (2u^2 cos^2 θ)] x^2

This is of the form y = Ax − Bx^2, which is the equation of a parabola. So the path of any projectile (ignoring air resistance) is always parabolic.

✓Final answer
$The path of projectile motion is a parabola.$
CBSE 2020Set ANNUAL1 mark
$Q. Answer in one word/one sentence: Which quantity is constant in projectile motion?$
›Reveal solutionSolution

Gravity gives the projectile a constant downward acceleration g, but no horizontal acceleration, so the horizontal velocity component vx stays fixed for the whole flight.

During projectile motion (ignoring air resistance), the only force acting is gravity, which acts vertically downward. This produces a constant downward acceleration g, changing the vertical velocity component continuously (vy = uy - gt). There is no horizontal force, so the horizontal acceleration is zero and the horizontal velocity component vx = ux remains unchanged throughout the motion.

✓Final answer
$The horizontal component of velocity (vx) is the quantity that remains constant in projectile motion.$

Q.An object thrown from a moving bus is an example of (A) Uniform circular motion (B) Rectilinear motion (C) Projectile motion (D) Motion in one dimension

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Q.An object thrown from a moving bus is an example of
(A) Uniform circular motion
(B) Rectilinear motion
(C) Projectile motion
(D) Motion in one dimension

Unlock everything free for 14 days