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

# Projectile motion (application)

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

Summary: Solving problems is an essential part of the understanding process.

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.

## Representative problems and their solutions

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

• Direction of motion on return
• Maximum height
• Equation of projectile motion
• Change in angles during motion
• Kinetic energy of a projectile
• Change in the direction of velocity vector

## Direction of motion on return

### Example 1

Problem : A projectile is thrown with a speed of 15 m/s making an angle 60° with horizontal. Find the acute angle, "α", that it makes with the vertical at the time of its return on the ground (consider g = 10 m / s 2 m / s 2 ).

Solution : The vertical component of velocity of the projectile at the return on the ground is equal in magnitude, but opposite in direction. On the other hand, horizontal component of velocity remains unaltered. The figure, here, shows the acute angle that the velocity vector makes with vertical.

The trajectory is symmetric about the vertical line passing through point of maximum height. From the figure, the acute angle with vertical is :

α = 90 0 - θ = 90 0 - 60 0 = 30 0 α = 90 0 - θ = 90 0 - 60 0 = 30 0

## Maximum height

### Example 2

Problem : Motion of a projectile is described in a coordinate system, where horizontal and vertical directions of the projectile correspond to x and y axes. The velocity of the projectile is 12i + 20j m/s at an elevation of 15 m from the point of projection. Find the maximum height attained by the projectile (consider g = 10 m / s 2 m / s 2 ).

Solution : Here, the vertical component of the velocity (20 m/s) is positive. It means that it is directed in positive y-direction and that the projectile is still ascending to reach the maximum height. The time to reach the maximum height is obtained using equation of motion in vertical direction :

v y = u y - g t v y = u y - g t

0 = 20 - 10 t 0 = 20 - 10 t t = 2 s t = 2 s

Now, the particle shall rise to a vertical displacement given by :

y = u y t - 1 2 g t 2 = 20 x 2 - 5 x 2 2 = 20 m y = u y t - 1 2 g t 2 = 20 x 2 - 5 x 2 2 = 20 m

The maximum height, as measured from the ground, is :

H = 15 + 20 = 35 m H = 15 + 20 = 35 m

## Equation of projectile motion

### Example 3

Problem : The equation of a projectile is given as :

y = 3 x - 1 2 g x 2 y = 3 x - 1 2 g x 2

Then, find the speed of the projection.

Solution : The general equation of projectile is :

y = x tan θ - g x 2 2 u 2 cos 2 θ y = x tan θ - g x 2 2 u 2 cos 2 θ

On the other hand, the given equation is :

y = 3 x - 1 2 g x 2 y = 3 x - 1 2 g x 2

Comparing two equations, we have :

tan θ = 3 θ = 60 0 tan θ = 3 θ = 60 0

Also,

u 2 cos 2 θ = 1 u 2 = 1 cos 2 θ u 2 = 1 cos 2 60 = 4 u = 2 m / s u 2 cos 2 θ = 1 u 2 = 1 cos 2 θ u 2 = 1 cos 2 60 = 4 u = 2 m / s

## Change in angles during motion

### Example 4

Problem : A projectile is projected at an angle 60° from the horizontal with a speed of ( 3 + 1 3 + 1 ) m/s. The time (in seconds) after which the inclination of the projectile with horizontal becomes 45° is :

Solution : Let "u" and "v" be the speed at the two specified angles. The initial components of velocities in horizontal and vertical directions are :

u x = u cos 60 0 u y = u sin 60 0 u x = u cos 60 0 u y = u sin 60 0

Similarly, the components of velocities, when projectile makes an angle 45 with horizontal, in horizontal and vertical directions are :

v x = v cos 45 0 v y = v sin 45 0 v x = v cos 45 0 v y = v sin 45 0

But, we know that horizontal component of velocity remains unaltered during motion. Hence,

v x = u x v cos 45 0 = u cos 60 0 v = u cos 60 0 cos 45 0 v x = u x v cos 45 0 = u cos 60 0 v = u cos 60 0 cos 45 0

Here, we know initial and final velocities in vertical direction. We can apply v = u +at in vertical direction to know the time as required :

v sin 45 0 = u + a t = u sin 60 0 - g t v cos 45 0 = u cos 60 0 t = u sin 60 0 - v sin 45 0 g v sin 45 0 = u + a t = u sin 60 0 - g t v cos 45 0 = u cos 60 0 t = u sin 60 0 - v sin 45 0 g

Substituting value of "v" in the equation, we have :

t = u sin 60 0 - u ( cos 60 0 cos 45 0 ) X sin 45 0 g t = u g ( sin 60 0 - cos 60 0 ) t = ( 3 + 1 ) 10 { ( 3 - 1 ) 2 ) t = 2 20 = 0.1 s t = u sin 60 0 - u ( cos 60 0 cos 45 0 ) X sin 45 0 g t = u g ( sin 60 0 - cos 60 0 ) t = ( 3 + 1 ) 10 { ( 3 - 1 ) 2 ) t = 2 20 = 0.1 s

## Kinetic energy of a projectile

### Example 5

Problem : A projectile is thrown with an angle θ from the horizontal with a kinetic energy of K Joule. Find the kinetic energy of the projectile (in Joule), when it reaches maximum height.

Solution : At the time of projection, the kinetic energy is given by :

K = 1 2 m u 2 K = 1 2 m u 2

At the maximum height, vertical component of the velocity is zero. On the other hand, horizontal component of the velocity of the particle does not change. Thus, the speed of the particle, at the maximum height, is equal to the magnitude of the horizontal component of velocity. Hence, speed of the projectile at maximum height is :

v = u cos θ v = u cos θ

The kinetic energy at the maximum height, therefore, is :

K = 1 2 m ( u cos θ ) 2 K = 1 2 m ( u cos θ ) 2

Substituting value of "u" from the expression of initial kinetic energy is :

K = m x 2 x K 2 m cos 2 θ K = K cos 2 θ K = m x 2 x K 2 m cos 2 θ K = K cos 2 θ

## Change in the direction of velocity vector

### Example 6

Problem : A projectile with a speed of “u” is thrown at an angle of “θ” with the horizontal. Find the speed (in m/s) of the projectile, when it is perpendicular to the direction of projection.

Solution : We need to visualize the direction of the projectile, when its direction is perpendicular to the direction of projection. Further, we may look to determine the direction of velocity in that situation.

The figure, here, shows the direction of velocity for the condition, when the direction of projectile is perpendicular to the direction of projection. From ΔOAB,

∠OBA = 180 0 - ( 90 0 + θ ) = 90 0 - θ ∠OBA = 180 0 - ( 90 0 + θ ) = 90 0 - θ

Thus, the acute angle between projectile and horizontal direction is 90- θ for the given condition. Now, in order to determine the speed, we use the fact that horizontal component of velocity does not change.

v cos ( 90 0 - θ ) = u cos θ v sin θ = u cos θ v = u cot θ v cos ( 90 0 - θ ) = u cos θ v sin θ = u cos θ v = u cot θ

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