Projectile On An Inclined Plane

Projectile motion on an inclined plane is a fascinating extension of classical mechanics that explores the behaviour of objects launched into the air at an angle relative to a sloped surface. Unlike standard projectile motion, where the ground is flat, the inclined plane adds complexity by introducing additional components of gravitational force and altering the trajectory. This type of motion is not just a theoretical concept but has real-world applications, such as in the design of ski slopes, the trajectory of a ball rolling down a hill, or even the path of water flowing down a roof. Understanding projectile motion on an inclined plane helps in predicting the path and final destination of objects in various engineering and natural scenarios, making it a vital concept in physics and applied sciences.

JEE Main 2025: Physics Formula | Study Materials | High Scoring Topics | Preparation Guide

JEE Main 2025: Syllabus | Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

NEET 2025: Syllabus | High Scoring Topics | PYQs

This Story also Contains

1. Projectile on an Inclined Plane
2. Important Terms
3. Solved Examples Based on Projectile on an Inclined Plane
4. Summary

Projectile on an Inclined Plane

Projectile motion on an inclined plane involves the study of an object's trajectory when it is launched at an angle on a surface that is itself tilted. This scenario differs from traditional projectile motion on flat ground, as the inclined plane introduces an extra dimension of complexity. The motion is influenced by both the angle of projection and the incline of the plane, which affects the range, maximum height, and time of flight.

Important Equations

$U=$ Speed of projection
$\alpha =$ The angle of projection above-inclined plane (measured from the horizontal line)
$\theta =$ The angle of projection above-inclined plane (measured from the inclined plane)
$\beta =$ The angle of inclination.

Initial Velocity (U)

Component along x or along inclined plane $=U_x=U\cos \theta$
Component along y or perpendicular to inclined plane $=U_y=U\mathrm{Sin}\theta$

Final velocity (V)

Component along x or along inclined plane $=V_x=U\cos \theta -(g\sin \beta )\cdot t$
Component along y or perpendicular to inclined plane $=V_y=U\sin \theta -(g\cos \beta )\cdot t$
and,

$V=\sqrt{V_x^2+V_y^2}$

Displacement (S)

Component along x or along inclined plane $S_x=U_{x}t+\frac{1}{2}{a}_x\cdot t^2$

Component along y or perpendicular to inclined plane $S_y=U_{y}t+\frac{1}{2}{a}_y\cdot t^2$

And $S=\sqrt{S_x^2+S_y^2}$

Acceleration (a)

Component along x or along inclined plane $=a_x=-g\sin \beta$
Component along y or perpendicular to inclined plane $=a_y=-g\cos \beta$
So $a=-g$

Important Terms

Time of flight

Formula

$T=\frac{2U\sin \theta }{g\cos \beta }$

Range along incline plane

Formula

$R=\frac{2u^2\cdot \sin (\alpha -\beta )\cdot \cos \alpha }{g{\cos }^2\beta }$

Recommended Topic Video

Solved Examples Based on Projectile on an Inclined Plane

Example 1: A plane surface is inclined making at angle $\theta$ with the horizontal. from the bottom of this inclined plane, a bullet is fired with velocity v. The maximum possible range of the bullet on the inclined plane is

1) $\frac{v^2}{g}$
2) $\frac{v^2}{g(1+\sin \theta )}$
3) $\frac{v^2}{g(1-\sin \theta )}$
4) $\frac{v^2}{g(1+\cos \theta )}$

Solution:

Projectile on an inclined plane

Important equations

U=Speed of projection

$\alpha =$ The angle of projection above-inclined plane (measured from the horizontal line)
$\theta =$ The angle of projection above-inclined plane (measured from the inclined plane)
$\beta =$ The angle of inclination.

Initial Velocity (U)

Component along x or along inclined plane $=U_x=U\cos \theta$
Component along y or perpendicular to inclined plane $=U_y=U\mathrm{Sin}\theta$

Final Velocity (V)

Component along x or along inclined plane $=V_x=U\cos \theta -(g\sin \beta )\cdot t$
Component along y or perpendicular to inclined plane $=V_y=U\sin \theta -(g\cos \beta )\cdot t$
and,

$V=\sqrt{V_x^2+V_y^2}$

Displacement (S)

Component along x or along inclined plane $=S_x=U_{x}t+\frac{1}{2}{a}_x\cdot t^2$
Component along y or perpendicular to inclined plane $=S_y=U_{y}t+\frac{1}{2}{a}_y\cdot t^2$
And $S=\sqrt{S_x^2+S_y^2}$

Acceleration (a)

Component along x or along inclined plane $a_x=-g\sin \beta$
Component along y or perpendicular to inclined plane $a_y=-g\cos \beta$
So $\mathrm{a}=-\mathrm{g}$
For maximum range $\sin (2\alpha -\theta {)}_{\text{should be the maximum}}$

$\begin{array}{rl}& \text{So for}(2\alpha -\theta )=\frac{x}{2}\\ & R=\frac{u^2}{g{\cos }^2\theta }[1-\sin \theta ]\\ & R=\frac{u^2}{g(1-{\sin }^2\theta )}[1-\sin \theta ]\\ & R=\frac{u^2}{g(1+{\sin }^2\theta )}\end{array}$

Hence, the answer is the option (2).

Example 2: A plane surface is inclined making an angle $\theta$ with the horizontal. from the bottom of this inclined plane, a bullet is fired with velocity v. The maximum possible range of the bullet on the inclined plane is

1) $\frac{v^2}{g}$
2) $\frac{v^2}{g(1+\sin \theta )}$
3) $\frac{v^2}{g(1-\sin \theta )}$
4) $\frac{v^2}{g(1+\cos \theta )}$

Solution:

For projectile on an inclined plane

The range on an inclined plane up to the plane is given
$\begin{array}{rl}& R=\frac{u^2}{g{\cos }^2\theta }\cdot 2\sin (\alpha -\theta )\cos \alpha \\ & \Rightarrow R=\frac{u^2}{g{\cos }^2\theta }[\sin (2\alpha -\theta )-\sin \theta ]\end{array}$
Where
$\alpha =$ The angle of projection above-inclined plane
(measured from the horizontal line)
$\theta =$ Angle of inclination.
So for the maximum range, $\sin (2\alpha -\theta )$ should be the maximum.

$\begin{array}{rl}& \text{so for}(2\alpha -\theta )=\frac{\pi }{2}\\ & R=\frac{u^2}{g{\cos }^2\theta }[1-\sin \theta ]\\ & R=\frac{u^2}{g(1-{\sin }^2\theta )}[1-\sin \theta ]\\ & R=\frac{u^2}{g(1+\sin \theta )}\end{array}$

Hence, the answer is the option (2).

Example 3: A projectile is launched from the foot of an inclined plane which makes an angle of 30 degrees with the horizontal. The projectile's initial velocity is 20 m/s at an angle of 45 degrees with the inclined plane. Neglecting air resistance, the time taken by the projectile to hit the inclined plane is closest to :

1) 1.0 s

2) 1.5 s

3) 2.0 s

4) 2.5 s

Solution:

The projectile's motion can be divided into two parts:

Horizontal motion and vertical motion.

In the horizontal direction, the projectile moves with a constant velocity of $20\cos (45)=14.14\mathrm{m}/\mathrm{s}$.

In the vertical direction, the projectile experiences a constant acceleration due to gravity of $9.8\mathrm{m}/{\mathrm{s}}^2$.

Let's consider the vertical motion of the projectile.

The initial vertical velocity of the projectile is $20\sin (45)=14.14\mathrm{m}/\mathrm{s}$.

The time taken by the projectile to hit the inclined plane can be found using the equation:

$y=v_{i}t+\frac{1}{2}a\times t^2$

where y is the vertical displacement of the projectile, is the initial vertical velocity of the projectile, is the acceleration due to gravity, and t is the time taken by the projectile to hit the inclined plane.

The vertical displacement of the projectile can be found using the angle of the inclined plane:

$y=x\tan 30$

where x is the horizontal displacement of the projectile.

The horizontal displacement of the projectile can be found using the time taken by the projectile to hit the inclined plane and the horizontal velocity of the projectile:

$x=14.14\times \cos (45)\times t$

Substituting these equations into the first equation, we get

$14.14\tan (30)t=14.14t\cos 45t+\frac{1}{29.8}\times t^2$

Simplifying and solving for t, we get:

$t\approx 1.5\mathrm{s}$

Hence, the answer is the option (2).

Example 4:

in the above case, what is the Component of Displacement along y or perpendicular to the inclined plane

1) $S_x=U_{x}t+\frac{1}{2}{a}_x\cdot t^2$
2) $S_x=2U_{x}t+\frac{1}{2}{a}_x\cdot t^2$
3) $S_x=U_{x}t+\frac{1}{2}{a}_x\cdot t^2$
4) $S_x=U_{x}t+\frac{1}{2}{a}_x\cdot t^2$

where

$\begin{array}{rl}& U_x=U\cos \frac{2}{3}\theta \\ & a_x=-g\sin \beta \end{array}$

Solution:
$S_x=U_{x}t+\frac{1}{2}{a}_x\cdot t^2$

where

$\begin{array}{rl}& U_x=U\cos \theta \\ & a_x=-g\sin \beta \end{array}$

Hence, the answer is the option (1).

Example 5: A plane surface is inclined making at angle $\theta$ with the horizontal. from the bottom of this inclined plane, a bullet is fired with velocity $v$. The maximum possible range of the bullet on the inclined plane is

1) $\frac{v^2}{g}$
2) $\frac{v^2}{g(1+\sin \theta )}$
3) $\frac{v^2}{g(1-\sin \theta )}$
4) $\frac{v^2}{g(1+\cos \theta )}$

Solution:

Projectile on an inclined plane

1. Important equations

U=Speed of projection

$\begin{array}{rl}& \alpha =\text{The angle of projection above-inclined plane (measured from the horizontal line)}\\ & \theta =\text{The angle of projection above-inclined plane (measured from the inclined plane)}\\ & \beta =\text{The angle of inclination.}\end{array}$

a) Initial Velocity(U)

Component along x or along inclined plane $=U_x=U\cos \theta$
Component along y or perpendicular to inclined plane $=U_y=U\mathrm{Sin}\theta$

b) Final velocity(V)

Component along x or along inclined plane $=V_x=U\cos \theta -(g\sin \beta )\cdot t$
Component along y or perpendicular to inclined plane $=V_y=U\sin \theta -(g\cos \beta )\cdot t$
and,

$V=\sqrt{V_x^2+V_y^2}$

c) Displacement(S)

Component along x or along inclined plane $=S_x=U_{x}t+\frac{1}{2}{a}_x\cdot t^2$ Component along y or perpendicular to inclined plane $=S_y=U_{y}t+\frac{1}{2}{a}_y\cdot t^2$

And $S=\sqrt{S_x^2+S_y^2}$

d) Acceleration(a)

xComponent along x or along inclined plane $=a_x=-g\sin \beta$
Component along y or perpendicular to inclined plane $=a_y=-g\cos \beta$
so $a=-g$
For maximum range $\sin (2\alpha -\theta )$ should be the maximum

So for $(2\alpha -\theta )=\frac{x}{2}$

$\begin{array}{rl}& R=\frac{u^2}{g{\cos }^2\theta }[1-\sin \theta ]\\ & R=\frac{u^2}{g(1-{\sin }^2\theta )}[1-\sin \theta ]\\ & R=\frac{u^2}{g(1+{\sin }^2\theta )}\end{array}$

Hence, the answer is the option (2).

Summary

The inclined plane projectile motion raises the most important concept because it extends our understanding of how objects move when launched onto surfaces that are not horizontal. This finds countless applications in skiing, engineering, sports, and education using both practical and theoretical aspects. Through the detailed study of this type of motion, it becomes possible to more accurately predict and optimize the behaviour of objects under numerous real-world conditions, thus assigning safer design and improved performance in various fields of human pursuit.