A particle is projected into the air with speed $u$ at an angle $\beta$ measured from an inclined plane. The plane itself is inclined at an angle $\alpha$ to the horizontal, so the direction of projection makes an angle $(\alpha + \beta)$ with the horizontal. The particle rises, then strikes the inclined surface again. Find: (a) an expression for the range along the inclined surface (the distance, measured along the plane, from the point of projection to the point where the particle lands on the plane); (b) the time of flight; (c) the value of $\beta$ for which this range along the plane is maximum.

A particle is projected into the air with speed u at an angle β measured from an inclined plane. The plane its

NCERT Exemplar · Q31

u

(a) an expression for the range along the inclined surface (the distance, measured along the plane, from the point of projection to the point where the particle lands on the plane);

(b) the time of flight;

\beta

for which this range along the plane is maximum.

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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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Choose axes along and perpendicular to the incline so that landing back on the plane simply means the perpendicular displacement returns to zero. This gives the time of flight directly; feeding it into the along-plane motion gives the range, and calculus (or a product-to-sum identity) gives the launch angle for maximum range.

Setting up tilted axes

Let $x$ run up the incline and $y$ run perpendicular to it. The initial velocity (speed $u$ at angle $\beta$ to the plane) resolves as

u_x = u\cos\beta, \qquad u_y = u\sin\beta.

Gravity $g$ (vertically down) resolves into these tilted axes as

g_x = -g\sin\alpha \ (\text{down the slope}), \qquad g_y = -g\cos\alpha \ (\text{into the plane}).

(b) Time of flight

The particle lands when its perpendicular displacement is again zero:

y = u_y t + \tfrac12 g_y t^2 = u\sin\beta\,t - \tfrac12 g\cos\alpha\,t^2 = 0.

Discarding $t=0$ ,

\boxed{\,T = \dfrac{2u\sin\beta}{g\cos\alpha}\,}.

(a) Range along the plane

The along-plane displacement at $t = T$ :

L = u_x T + \tfrac12 g_x T^2 = u\cos\beta\,T - \tfrac12 g\sin\alpha\,T^2.

Insert $T = \dfrac{2u\sin\beta}{g\cos\alpha}$ :

L = \frac{2u^2\sin\beta\cos\beta}{g\cos\alpha} - \frac{2u^2\sin^2\beta\,\sin\alpha}{g\cos^2\alpha} = \frac{2u^2\sin\beta}{g\cos^2\alpha}\big[\cos\beta\cos\alpha - \sin\beta\sin\alpha\big].

The bracket is $\cos(\alpha+\beta)$ , so

\boxed{\,L = \dfrac{2u^2\sin\beta\,\cos(\alpha+\beta)}{g\cos^2\alpha}\,}.

(c) Angle for maximum range

Only the factor $f(\beta) = \sin\beta\,\cos(\alpha+\beta)$ depends on $\beta$ . Using the product-to-sum identity, …

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Concept: Intersect the Ordinary (Un-Rotated) Parabola With the Incline's Line, Instead of Rotating the Axes

Method: Standard Horizontal/Vertical Axes + the Incline as a Line $y=x\tan\alpha$

The direct method rotates the coordinate axes to align with the incline. This method keeps the ordinary horizontal ( $x$ ) and vertical ( $y$ ) axes throughout, writes the incline simply as the line $y=x\tan\alpha$ , and finds the landing point as the second intersection of the unrotated projectile parabola with that line.

Steps

Write the launch velocity components in the ordinary horizontal/vertical axes. The particle leaves at angle $(\alpha+\beta)$ to the horizontal with speed $u$ :

v_x = u\cos(\alpha+\beta), \qquad v_y = u\sin(\alpha+\beta).

Write the ordinary projectile equations of motion (gravity is simply $-g$ vertically, as always):

x(t) = u\cos(\alpha+\beta)\,t, \qquad y(t) = u\sin(\alpha+\beta)\,t - \tfrac12 g t^2.

Write the incline as the line $y = x\tan\alpha$ and substitute:

u\sin(\alpha+\beta)\,t - \tfrac12 g t^2 = u\cos(\alpha+\beta)\tan\alpha\;t.

Divide through by $t$ (excluding $t=0$ ) and solve for $t$ :

u\big[\sin(\alpha+\beta) - \cos(\alpha+\beta)\tan\alpha\big] = \tfrac12 g t.

Combine the bracket over $\cos\alpha$ using the sine-subtraction identity:

\sin(\alpha+\beta) - \cos(\alpha+\beta)\tan\alpha = \frac{\sin(\alpha+\beta)\cos\alpha - \cos(\alpha+\beta)\sin\alpha}{\cos\alpha} = \frac{\sin\beta}{\cos\alpha}.

So $T = \dfrac{2u\sin\beta}{g\cos\alpha}$ -- identical to the tilted-axis result, obtained without ever rotating anything.

Convert the horizontal landing distance to a distance along the slope using $x=L\cos\alpha$ : $x(T) = u\cos(\alpha+\beta)\cdot\frac{2u\sin\beta}{g\cos\alpha}, \qquad L = \frac{x(T)}{\cos\alpha} = \frac{2u^2\sin\beta\cos(\alpha+\beta)}{g\cos^2\alpha}.$ …

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Setting up tilted axes

(b) Time of flight

(a) Range along the plane

(c) Angle for maximum range

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