# OpenStax-CNX

You are here: Home » Content » Physics for K-12 » Projectile motion on an incline (application)

• Why yet another course in physics?
• What is physics?

### Recently Viewed

This feature requires Javascript to be enabled.

Inside Collection (Course):

Course by: Sunil Kumar Singh. E-mail the author

# Projectile motion on an incline (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

## Hints for solving problems

Problems based on projectile motion over an incline are slightly difficult. The analysis is complicated mainly because there are multitudes of approaches available. First there is issue of coordinates, then we might face the conflict to either use derived formula or analyze motion independently in component directions and so on. We also need to handle motion up and down the incline in an appropriate manner. However, solutions get easier if we have the insight into the working with new set of coordinate system and develop ability to assign appropriate values of accelerations, angles and component velocities etc.

Here, we present a simple set of guidelines in a very general way :

1: Analyze motion independently along the selected coordinates. Avoid using derived formula to the extent possible.

2: Make note of information given in the question like angles etc., which might render certain component of velocity zero in certain direction.

3: If range of the projectile is given, we may try the trigonometric ratio of the incline itself to get the answer.

4: If we use coordinate system along incline and in the direction perpendicular to it, then always remember that component motion along both incline and in the direction perpendicular to it are accelerated motions. Ensure that we use appropriate components of acceleration in the equations of motion.

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the concept of projectile motion on an incline. The questions are categorized in terms of the characterizing features of the subject matter :

• Range of the flight
• Angle of projection
• Final Speed of the projectile
• Elastic collision with the incline
• Projectile motion on double inclines

## Range of the flight

### Example 1

Problem : A projectile is thrown with a speed "u" at an angle 60° over an incline of 30°. If the time of flight of the projectile is “T”, then find the range of the flight.

Solution : We can see here that time of flight is already given. We can find range considering projectile motion in the coordinates of horizontal and vertical axes. The range of the projectile “R” is obtained by using trigonometric ratio in triangle OAB. The range is related to horizontal base “OB” as :

cos 30 0 = O B O A cos 30 0 = O B O A

R = O A = O B cos 30 0 = O B sec 30 0 R = O A = O B cos 30 0 = O B sec 30 0

Now, we can find OB by considering motion in horizontal direction :

R = O B = u x T = u cos 60 0 T = u T 2 R = O B = u x T = u cos 60 0 T = u T 2

Thus, the range of the projectile, OA, is :

R = O A = u T sec 30 0 2 = u T 3 R = O A = u T sec 30 0 2 = u T 3

## Angle of projection

### Example 2

Problem : A particle is projected from the foot of an incline of angle “30°” at a certain velocity so that it strikes the incline normally. Find the angle of projection (θ ) as measured from the horizontal.

Solution : Here, projectile hits the incline normally. It means that component of velocity along the incline is zero. We should remember that the motion along the incline is not uniform motion, but a decelerated motion. In order to take advantage of the fact that final component velocity along incline is zero, we consider motion in a coordinate system along the incline and along a direction perpendicular to it.

v x = u x + a x T v x = u x + a x T

0 = u cos θ 30 0 g sin 30 0 T 0 = u cos θ 30 0 g sin 30 0 T

T = u cos θ 30 0 g sin 30 0 T = u cos θ 30 0 g sin 30 0

Now, time of flight is also given by the formulae :

T = 2 u sin θ 30 0 g cos 30 0 T = 2 u sin θ 30 0 g cos 30 0

Equating two expressions for time of flight, we have :

2 u sin θ 30 0 g cos 30 0 = u cos θ 30 0 g sin 30 0 2 u sin θ 30 0 g cos 30 0 = u cos θ 30 0 g sin 30 0

tan θ 30 0 = 1 2 tan 30 0 = 3 2 tan θ 30 0 = 1 2 tan 30 0 = 3 2

θ = 30 0 + tan - 1 3 2 θ = 30 0 + tan - 1 3 2

## Final speed of the projectile

### Example 3

Problem : A ball is projected on an incline of 30° from its base with a speed 20 m/s, making an angle 60° from the horizontal. Find the speed with which the ball hits the incline.

Solution : We analyze this problem in the coordinates along the incline (x-axis) and in the direction perpendicular to the incline (y-axis). In order to find the speed at the end of flight, we need to find the component velocities in “x” and “y” directions.

The velocity in y-direction can be determined making use of the fact that a ball under constant acceleration like gravity returns to the ground with the same speed, but in opposite direction. The component of velocity in y-direction at the end of the journey, therefore, is :

v y = u y = 20 sin 30 0 = 20 X 1 2 = - 10 m / s v y = u y = 20 sin 30 0 = 20 X 1 2 = - 10 m / s

Now, we should attempt to find the component of velocity in x-direction. We should, however, recall that motion in x-direction is not a uniform motion, but has deceleration of “-gsinα”. Using equation of motion in x-direction, we have :

v x = u x + a x T v x = u x + a x T

Putting values,

v x = u cos 30 0 g sin 30 0 T = 20 X 3 2 10 X 1 2 X T = 10 3 5 T v x = u cos 30 0 g sin 30 0 T = 20 X 3 2 10 X 1 2 X T = 10 3 5 T

Clearly, we need to know time of flight to know the component of velocity in x-direction. The time of flight is given by :

T = 2 u y a y = 2 u sin 30 0 g cos 30 0 = 2 X 20 3 X 10 = 4 3 T = 2 u y a y = 2 u sin 30 0 g cos 30 0 = 2 X 20 3 X 10 = 4 3

Hence, component of velocity in x-direction is :

v x = 10 3 5 T = 10 3 5 X 4 3 v x = 10 3 5 T = 10 3 5 X 4 3

v x = 30 - 20 3 = 10 3 v x = 30 - 20 3 = 10 3

The speed of the projectile is equal to resultant of components :

v = { - 10 2 + 10 3 2 } v = { - 10 2 + 10 3 2 }

v = 400 3 = 20 3 m / s v = 400 3 = 20 3 m / s

## Elastic collision with the incline

### Example 4

Problem : A ball falls through a height “H” and impacts an incline elastically. Find the time of flight between first and second impact of the projectile on the incline.

Solution : The ball falls vertically through a distance "H". We can get the value of initial speed by considering free fall of the ball before impacting incline.

0 = u 2 2 g H 0 = u 2 2 g H

u = 2 g H u = 2 g H

In order to answer this question, we need to identify the velocity with which projectile rebounds. Since impact is considered elastic, the projectile is rebounded without any loss of speed. The projectile is rebounded such that angle of incidence i.e. the angle with the normal is equal to angle of reflection.

The components of velocity along the incline and perpendicular to it are shown in the figure. The motion of ball, thereafter, is same as that of a projectile over an incline. Here, we shall analyze motion in y-direction (normal to the incline) to find the time of flight. We note that the net displacement between two strikes is zero in y – direction.

Applying equation of motion

y = u y T + 1 2 a y T 2 y = u y T + 1 2 a y T 2

0 = u cos α T 1 2 g cos α T 2 0 = u cos α T 1 2 g cos α T 2

T = 0 T = 0

Or

T = 2 u cos α g cos α = 2 u g T = 2 u cos α g cos α = 2 u g

Putting value of initial speed in the equation of time of flight, we have :

T = 8 H g T = 8 H g

It should be noted here that we can find the time of flight also by using standard formula of time of flight for projectile motion down the incline. The time of flight for projection down an incline is given as :

T = 2 u sin θ + α g cos α T = 2 u sin θ + α g cos α

We need to be careful while appropriating angles in the above expression. It may be recalled that all angles are measured from the horizontal. We redraw the figure to denote the value of angle of projection “θ” from the horizon.

T = 2 u sin 90 0 - 2 α + α g cos α = 2 u sin 90 0 - α g cos α T = 2 u sin 90 0 - 2 α + α g cos α = 2 u sin 90 0 - α g cos α

T = 2 u cos α g cos α = 2 u g = 8 H g T = 2 u cos α g cos α = 2 u g = 8 H g

## Projectile motion on two inclines

### Example 5

Problem : Two incline plane of angles 30° and 60° are placed touching each other at the base as shown in the figure. A projectile is projected at right angle with a speed of 10√3 m/s from point “P” and hits the other incline at point “Q” normally. Find the linear distance between PQ.

Solution : We notice here OPQ forms a right angle triangle at “O”. The linear distance, “PQ” is related as :

P Q 2 = O P 2 + O Q 2 P Q 2 = O P 2 + O Q 2

In order to find “PQ”, we need to know “OP” and “OQ”. We can find “OP”, considering motion in y-direction.

y = O P = u y T + 1 2 a y T 2 = 0 + 1 2 a y T 2 = 1 2 a y T 2 y = O P = u y T + 1 2 a y T 2 = 0 + 1 2 a y T 2 = 1 2 a y T 2

Similarly,

x = O Q = u x T + 1 2 a x T 2 = 10 3 T + 1 2 a x T 2 x = O Q = u x T + 1 2 a x T 2 = 10 3 T + 1 2 a x T 2

In order to evaluate these two relations, we need to find components of accelerations and time of flight.

Considering first incline, we have :

a x = - g cos 30 0 = - 10 X 3 2 = - 5 3 m / s 2 a x = - g cos 30 0 = - 10 X 3 2 = - 5 3 m / s 2

a y = - g sin 30 0 = - 10 X 1 2 = - 5 m / s 2 a y = - g sin 30 0 = - 10 X 1 2 = - 5 m / s 2

In order to find the time of flight, we can further use the fact that the component of velocity in x-direction i.e. along the second incline is zero. This, in turn, suggests that we can analyze motion in x-direction to obtain time of flight.

In x-direction,

v x = u x + a x T v x = u x + a x T

0 = u x + a x T 0 = u x + a x T

T = - u x a x T = - u x a x

Putting values in the equation and solving, we have :

T = - 10 3 5 3 = 2 s T = - 10 3 5 3 = 2 s

Now, we can evaluate “x” and “y” displacements as :

y = O P = 1 2 a y T 2 = - 1 2 X 5 X 2 2 = - 10 m y = O P = 1 2 a y T 2 = - 1 2 X 5 X 2 2 = - 10 m

x = O Q = 10 3 x 2 + 1 2 - 5 3 X 2 2 x = O Q = 10 3 x 2 + 1 2 - 5 3 X 2 2

x = O Q = 10 3 x = O Q = 10 3

Considering positive values, the linear distance, “PQ” is given as :

P Q = O P 2 + O Q 2 = { 10 2 + 10 3 2 } = 20 m P Q = O P 2 + O Q 2 = { 10 2 + 10 3 2 } = 20 m

## Content actions

PDF | EPUB (?)

### What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

#### Collection to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

#### Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks