Showing posts with label Two Dimensional Motion. Show all posts
Showing posts with label Two Dimensional Motion. Show all posts

Horizontal Projectile and its Velocity Video Lesson

A body  projected from a certain height horizontally with some initial velocity is called horizontal projectile. It has only initial velocity  along the horizontal direction but it has no initial velocity along the vertical direction. 

 But as the body start moving, acceleration due to gravity starts acting, velocity  of the body in the horizontal direction remains same as there is no acceleration due to gravity remains same. But velocity component of the body along the Y direction starts increasing. We can find the displacement along the horizontal and vertical directions using the basic equations of motion and we can show that the path is parabola as shown in the video lesson below.

 Path of horizontal Projectile is parabola


Velocity of horizontal projectile

The horizontal projectile has initial velocity of projection in the horizontal direction but it has no initial velocity along the Y direction. The velocity of projectile in the horizontal direction remains same as there is no acceleration in that direction and the component of the velocity along the Y direction starts increasing. We can find the final velocity of projectile at any instant is the vector sum of horizontal and vertical components of velocity. The final velocity has a certain direction and we can find the direction as shown in the video lecture below.


Angle of projectile of oblique projectile where range and maximum heights are same 

A oblique projectile is a body that is projected from the horizontal direction with a certain angle. The range of the projectile is the maximum horizontal distance travelled by a projectile body and the maximum height is the maximum vertical distance that the projectile has travelled. If these two has to he the same, we need to equate the mathematical equations of both of them and we can find the corresponding angle of projection can be measured as shown in the diagram below. 



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Range and Maximum height of Projectile Motion Video Lesson

Range of the projectile is the maximum horizontal distance travelled by a projectile during its time of flight. By the time the end of time of flight, the vertical displacement of the projectile is zero but its initial and final positions along the horizontal direction are different. We can find the horizontal displacement of a projectile as the product of velocity of projectile along the horizontal direction and time taken for the journey. The time taken to complete projectile motion is called time of flight. The horizontal displacement that the projectile has during the time of flight is the maximum horizontal displacement and it is called range of the projectile. We can find the range of the projectile as shown in the video lesson below.



It is found that the range of a projectile depends on the initial velocity of projection and angle of projection. So it is clear that the range of a projectile is maximum for an angle of projection 45 degree. It is the reason due to which people try to project any body with an angle close to 45 degree when they want to get maximum horizontal distance like disc and javelin throw.

Range is same for two angles of projection

It can be proved that the range of the projectile is same for complimentary angels of projection. The heights that they reach is different but the maximum horizontal range that both of them takes for complimentary angles are same. It is shown in video lectures as shown in the video below.


Maximum height of projectile

Maximum height of the projectile is the maximum vertical displacement that a projectile has during its journey. We can measure it using the equation of motion of one dimension itself. The projectile has initial velocity component along the vertical direction. As it is moving up against the gravity, its velocity keeps on decreasing and the point at which the vertical component of velocity becomes zero is called maximum direction. We can derive the equation in the video lesson as shown in the diagram below.



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Motion of body in a Plane and Projectile motion Video Lesson

Any body moving in a plane where it has motion along horizontal and vertical directions, then the motion is called motion in a plane and it is also called two dimensional motion or projectile motion. For a body to have projectile motion, it shall be projected from the ground with an angle other than ninety  degree with a certain initial velocity.The body has sentimentally velocity along both the directions.  To find the velocity along a given direction, we need to resolve the initial velocity has to be resolved into components. Some component is along the horizontal axis and as there is no gravity along that direction and hence velocity component along the horizontal direction remains constant. The velocity along the vertical directions changes as acceleration due to gravity acts against initial vertical direction and hence the velocity keep changing along that direction. We can find the displacement along the horizontal and vertical directions. It is explained in the video lesson as shown in the video below.


Path of projectile is parabola

When a body is projected with an angle to the horizontal, it has displacement along both horizontal and vertical directions. We can find them using the equation of motion as shown in the diagram below. The displacement along X direction is independent of acceleration due to gravity as it is acting only along Y direction. We can find the time taken to have horizontal direction in terms of initial velocity as shown in the video lesson below. We can find the displacement along Y direction and we need to substitute the time from the first case and hence we will get a vertical displacement format as mathematical parabola.


Time of flight of Projectile

The time taken by the projectile to reach the ground from the point of projection is called time of flight. We need to know that by the time of end of journey, it comes to some other point on the ground and hence it has some horizontal displacement. But its initial and final position with respect to the Y direction and hence the vertical displacement by the end of time of flight along the vertical direction is zero. Taking this into consideration, we can find the time of flight as  shown in the video lesson below.


Velocity of Projectile at any instant

The projectile has some initial velocity and it can be resolved into components along horizontal and vertical direction. As the time progresses, the total velocity of the projectile changes. The horizontal component of velocity of the projectile remains constant but the vertical component of the velocity keeps on changing. To find the final velocity, we need to add final velocity components of X and Y directions and it can be done as shown in the video lesson below.


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Motion in One and two Dimensions Problems with Solutions Fourteen

We are solving problems on one and two dimensional motion in this series of chapters. Here we are dealing with motion of a body moving along only one dimension and also we are studying the motion of a body in a plane where it has motion both along X and Y axis simultaneously.

In the given problem a body is dropped from a certain height and it strikes the inclined plane at a height h above the ground. As a result of impact, velocity of the body becomes horizontal. We need to measure  that it will take maximum time to reach the ground basing on what condition. The problem is  as shown in the diagram below.


Solution

It is given that the body is dropped from a certain height and hence its initial velocity is zero. It has fallen only some distance before it actually strikes the ground. That can be found as the difference between the total height from where it  is dropped and the location where it hit the ground.

Thus we can measure the time of fall in both the cases using the equations of motion. We need to measure the total maximum journey time. For the parameter to be maximum, its differentiation function with height has to become zero as a mathematical rule. 

By simplifying it, we can find that relation as shown in the diagram below.


Problem

It is given in the problem that a projectile is projected with a velocity and a known angle of projection. It is given that the trajectory grazes the vertices of the triangle as shown in the diagram. We need to know the relation between the angles with the horizontal.



Solution

The diagram for the problem data is as shown in the figure. We can express the angles with the horizontal in terms of horizontal and vertical displacement as shown in the diagram below.

We also know that the vertical displacement can be expressed in terms of horizontal displacement and angle of projection in terms of range as shown further.

By comparing this two equations, we can find the relation as shown below.


Problem

Two cannons are installed at the top of a cliff of 10 meter and they fire a shot with a known speed at some interval of time. The first cannon is fired with an angle and the other one is done horizontally. We need to know where do the two cannons collide with each other. The problem is as shown in the diagram below.


Solution

Both of them were fired from a certain height. But one horizontally and the other with an angle. By substituting the given data, we can simplify the equations as shown in the diagram below.



Problem

It is given in the problem that a particle is projected from the ground with a initial speed at angle of projection with the horizontal. We need to measure the average velocity of the projection between its point of projection and the highest point of the trajectory. The problem is as shown in the diagram below.


Solution

We know that the average velocity is the ratio of total displacement of a body in the total time of the journey. We can find the total displacement of the journey as a vector sum of horizontal and vertial displacements and the time of journy is half of time of flight.By substituting that data, we can  solve the problem as shown in the diagram below.



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Motion in One Dimension Problems with Solutions Thirteen

We are solving series of problems in one dimensional motion. It is given in the problem that a point moves in a straightline under a retardation as given in the problem where k is a constant.Acceleration is defined as rate of change of velocity. If a body has acceleration means, its velocity is increasing with time. If it is deceasing with  time, it is called retardation.

It is given that the particle has initial velocity. We need to know the distance covered by it in the given time. The problem is as shown in the diagram below.


Solution

We know that retardation is a negative acceleration. Here in this case the velocity of the body is decreasing with time. To indicate it, a negative sign is taken into consideration. We can rearrange the velocity  in terms of time as shown in the diagram below. As we know the part of velocity, to get the total velocity by integrating it in the given limits. By using rules of integration, we can the value of final velocity after applying the limits of lower and upper.

We need displacement and hence write the velocity as the rate of change of displacement. We can write the equation for small displacement and by integrating it, we can get the displacement of the particle as shown in the diagram below.


Problem

In the arrangement shown in the figure and instantaneous velocities of two masses, we need to know the angle between the vertical as shown in the diagram below. 


Solution

Let the vertical distance between pulley and the surface is X. Let the horizontal distance is 2a. It is given that the length of the string is constant and hence it can be found in the given terms as shown in the diagram below. As the equation is constant, its differentiation with time is zero.

By applying that condition, we can the relation between two velocities as shown. Basing on that we can find the angle and the detailed solutions is given below.


Problem

It is given in the problem that a ball is thrown vertically upward with a known velocity. While going upward and again while coming back after reaching its maximum height, the ball came back to the same point and the time interval between them is given to us in the problem. We need to measure the initial velocity of the ball. The problem is as shown in the diagram below.


Solution

Let the height that is crossed twice is and the intervals are given in the problem as shown in the diagram. Thus we can equate the height equation both in terms of the first time and the second time. By solving these two  equations, we can solve the problem as shown in the diagram below.


Problem

It is given in the problem that a body is projected vertically and different points are given in the journey. If the body is released from position A and we need to know the time of descents ratio need to be found.


Solution

We know that the time intervals for each part is same and hence to reach each point, the time taken in multiples of that interval as shown in the diagram below. We need to use the second equation of motion and solve the problem as shown in the diagram below.


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Motion in One Dimension Problems with Solutions Twelve

We are interested in solving the problems based on displacement,velocity and acceleration. They are all belongs to one dimensional motion. We know that the velocity is the rate of change of displacement and acceleration is the rage of change of velocity.

In the first problem it is given that position of the particle is given in the XY plane and it is given in terms of time and trigonometric function. We need to know the path of the particle and the problem is as shown below.


Solution

Position vector is given to us in terms of both X and Y components. By multiplying with I unit vector we identify it as the X component. Similarly by multiplying with J, we can identify the component as Y component. J is a unit vector along Y axis. Unit vector is a vector with specific direction and has magnitude of only one unit.

Using trigonometric rule, we can find the relation between X and Y as shown in the diagram below. The path is a circle as the equation supports that mathematically.


Problem

A particle moves according to the equation given in the problem with some constants. We need to find the velocity of the particle in terms of time and we can assume that the body starts from rest. The problem is as shown in the diagram below.


Problem

Rate of change of velocity is given in the problem and we can rearrange the terms to get the component of velocity. But we need the total velocity and to get that we shall integrate the component of velocity with certain given limits. By simplifying the equation further, we can solve the problem as shown in the diagram below.


Problem

A particle moves in a straight line and the position of the particle is given to us in terms of time as shown in the diagram below. We need to know how the velocity and acceleration of the particle changes with respect to time and we need to check with the options given in the diagram below.


Solution

What is given in the problem is in terms of displacement and to get the velocity we need to differentiate the equation. It can be found further that after two seconds velocity of the particle and its displacement becomes zero.

By differentiating the velocity further we can get acceleration as shown below. It can be concluded basing on the equations that if time is less than three seconds, acceleration could be negative.


Problem

It is given in the problem that after the engine is switched off, the boat goes for retardation and it is given as shown in the diagram below. We need to know the velocity of the particle after a specified time.


Solution

By rearranging the terms, we can get a part of velocity. To get the total velocity in the given limits, we need to integrate the equation. Thus we get the velocity of the particle at a given time as shown in the diagram below.



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Motion in One Dimension Problems with Solutions Ten

We would like to solve a problem basing on relative velocity concept. There are two particles separated from a origin and the two positions are perpendicular to each other. They are moving with a known velocity. We need to know the nearest distance between them. The problem is as  shown in the diagram below.


Solution

Let us consider that A and B are the two bodies in the given problem and they are located perpendicular to the origin. Let A has travelled for one second and hence it covers three meter in that time. As it is given that it is initially at ten meter from the origin, after one second it is at the distance of seven meter.

Simultaneously the body B has moved a distance four meter from the origin. So we can find the shortest distance between the two bodies as shown in the diagram below.


Problem

In this one dimensional motion problem, the position of the body is given in terms of time and constants as shown in the diagram below.We need to know the particular value of the time at which acceleration of the body is zero.

Solution

It is given that displacement in terms of time.By differentiating displacement once with respect to time we can velocity and by differentiating velocity with respect to time we get acceleration.We shall equate acceleration to zero as per given problem to get the time required in the problem.

By using the basic rule of differentiation, we can solve the problem as shown in the diagram below.


Problem

This problem is also a similar problem to the above. We need to know what is the displacement of the particle at the instant when the velocity of the body is zero.


Solution

The given equation is in terms of displacement and by rearranging them we can get the displacement equation in terms of time as shown in the diagram below.

By differentiating the displacement once with respect to time we get velocity. As per the condition given in the problem, we need to equate it to zero. Thus it is found that velocity of the body becomes zero after three seconds. By substituting that value in the displacement equation, we can get the displacement as  shown in the diagram below.


Problem

This problem is also about the relation between displacement and time with some constants involved. The problem is as shown in the diagram below. We need to find the displacement after  a certain time to verify the first option.

We need to check the velocity to verify the second option. The third option is regarding the new new displacement of the particle.


Solution

By differentiating the displacement once with respect to time as shown in the diagram below, we can get the velocity of the particle. By differentiating once again, we can get acceleration of the same particle. By substituting the given condition of the time, we can solve the problem as shown in the diagram below.


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