A particle falls vertically and strikes a plane surface that is inclined at an angle $\theta$ to the horizontal. At the moment of impact its speed is $v_0$ (directed vertically downward), and it rebounds elastically from the plane (so the speed just after the bounce is again $v_0$, with the velocity component perpendicular to the surface reversed and the component along the surface unchanged). Find the distance, measured along the inclined plane, from the first point of impact to the point where the particle strikes the plane a second time.

A particle falls vertically and strikes a plane surface that is inclined at an angle θ to the horizontal. At t

NCERT Exemplar · Q32

\theta

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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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After the bounce, work in axes along and perpendicular to the incline. The perpendicular launch component $v_0\cos\theta$ and the perpendicular gravity component $g\cos\theta$ set the flight time to $2v_0/g$ ; the along-plane motion, driven by $v_0\sin\theta$ plus the $g\sin\theta$ pull down the slope, then carries the particle a distance $4v_0^2\sin\theta/g$ .

Resolving the impact velocity

The particle arrives moving straight down with speed $v_0$ . The normal to the incline makes angle $\theta$ with the vertical, so relative to the surface the incoming velocity splits into

\text{along the plane (down-slope): } v_0\sin\theta, \qquad \text{perpendicular (into the plane): } v_0\cos\theta.

An elastic bounce off the smooth surface reverses the perpendicular component and leaves the parallel one untouched. So immediately after rebound:

u_x = v_0\sin\theta \ (\text{down the slope}), \qquad u_y = v_0\cos\theta \ (\text{away from the plane}).

Tilted axes and gravity components

With $x$ down the incline and $y$ perpendicular outward, gravity resolves as

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

Time of flight (return to the plane)

Set the perpendicular displacement back to zero: …

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Concept: Use the Vector Reflection Formula for the Rebound, Instead of Resolving Into Along/Perpendicular Components

Method: $\vec v_{\text{out}} = \vec v_{\text{in}} - 2(\vec v_{\text{in}}\cdot\hat n)\hat n$ (the Mirror-Reflection Formula)

An elastic bounce off a rigid surface obeys exactly the same rule light obeys reflecting off a mirror: reverse the velocity component along the surface normal, keep the tangential component. Applying the general vector reflection formula directly gives the rebound velocity in one step.

Steps

Set up ordinary horizontal ( $x$ ) / vertical ( $y$ ) axes, and write the incline's outward unit normal (surface descending to the right at angle $\theta$ ): $\hat n = (\sin\theta,\cos\theta)$ .
Write the incoming velocity (straight down, speed $v_0$ ): $\vec v_{\text{in}} = (0,-v_0)$ .
Apply the reflection formula directly:

\vec v_{\text{in}}\cdot\hat n = (0)(\sin\theta)+(-v_0)(\cos\theta) = -v_0\cos\theta.

\vec v_{\text{out}} = \vec v_{\text{in}} - 2(\vec v_{\text{in}}\cdot\hat n)\hat n = (0,-v_0) + 2v_0\cos\theta(\sin\theta,\cos\theta) = \big(v_0\sin2\theta,\ v_0\cos2\theta\big).

Notice $|\vec v_{\text{out}}|=v_0$ automatically -- the reflection formula preserves speed on its own, confirming "elastic" without a separate energy check.

Find the second landing point using the ordinary parabola and the incline line $y=-x\tan\theta$ :

v_0\cos2\theta\,t - \tfrac12gt^2 = -v_0\sin2\theta\tan\theta\,t.

Dividing by $t$ and combining over $\cos\theta$ using the cosine-subtraction identity:

$v_0\big[\cos2\theta+\sin2\theta\tan\theta\big] = \tfrac12gt \;\Rightarrow\; \frac{\cos2\theta\cos\theta+\sin2\theta\sin\theta}{\cos\theta} = \frac{\cos(2\theta-\theta)}{\cos\theta} = 1 \;\Rightarrow\; T=\frac{2v_0}{g}.$ …

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Resolving the impact velocity

Tilted axes and gravity components

Time of flight (return to the plane)

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