Showing posts with label Projectile. Show all posts
Showing posts with label Projectile. 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 Dimension Problems with Solutions Six

One dimensional motion is the study of the motion of a body only along one direction. In the given problem, a boy sees a ball going up and then down through the window of known height. The total time of visibility the window is given to us in the problem. We need to measure the height above the window. The problem is as shown in the diagram below.

Equations of motion help us in understanding the motion of the one dimensional body. If any three parameters of them are known, by using he appropriate relation, we can find the forth physical quantity.




Solution

As the total time of the journey is one second the upward or downward journey takes half of that time and that is half second. As it is all ready  crossing a certain height h before window, by the time it reaches the window, it will acquire a velocity and we can find that velocity using the third equation of motion.

By substituting that data in the second equation of the motion, we can measure the height above the window as shown in the diagram below.


Problem

In this problem, we are dealing with a body thrown vertically upward with a known velocity. While it is crossing the top of the tower, a man tries to catch it and failed to catch it and it has gone  over crossed above him. Any way in the return journey after three seconds, the ball reached the same person. The boy is able to catch it in the return journey and we are here to measure the height of the tower where that man is standing. The problem is as shown in the diagram below.



Solution

As the velocity of vertical projection is given to us, we can measure the time of ascent. The man is at the maximum height of the tower and he tries to catch it. It is obvious that the tower is at a height less than the maximum height that the stone can travel. He failed to catch it there and it has gone above him further. It is given that it returned to the same place after three seconds. It means it has travelled up further 1.5 second and started returning back. Thus the time for which it has crossed the tower is only one second. By substituting that data in the height of the tower formula, we can find the height of the tower as shown in the diagram below.



Problem

In the given  problem, two stones are thrown vertically upward with the same velocity. But the second stone is thrown after three seconds of the first stone. We need to measure where do these two stones are going to meet ?
Solution

As the bodies are moving against the gravity, we need to treat the acceleration due to gravity as negative. As the times are different by three seconds, we need to take the second stone time as first stone time and less than three seconds as shown in the diagram below. But the height travelled by them has to be same so that they can meet at a certain height. By equating these equations, we can  get the time after which the two stones are going to meet and by further substituting that time data in any one of the height equation, we can measure the height of the meeting point as shown in the diagram below.


Problem

In the given problem, a velocity and time graph is given. We need to measure the ratio of distance travelled by the body in the first two seconds when compared with the total distance travelled by the body in the total time of the journey.



Solution

In the velocity and time graph, we can find the distance in the two seconds can be found as the area between the two seconds. Similarly we can also  find the the total distance as the total area of the graph in the total time as shown in the diagram below.


Problem

In this problem, a ball is thrown vertically up from the ground with a known velocity. At the same time ,another body is thrown down from a known height.It is given in the problem that after some time the two bodies got the collision and here we need to find the velocity of the each body at the time of meeting .


Solution

For the vertically thrown body, acceleration due to gravity shall be treated as negative and for the vertically thrown down body it is positive. For the second body its initial velocity is zero. Equating the equations of both the cases, we can find the specific time after which the stones are going to meet. Again by using that time value, we can find velocities of the bodies using the first equation of the motion as shown in the diagram below.




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Horizontal Projectile, Applications Problems with Solutions

A body projected horizontally from a certain height with an initial horizontal velocity can be called as a horizontal projectile. Its initial velocity along the vertical direction is zero and it possess only horizontal velocity at the beginning. As the time progresses, due to the impact of the gravity, it acquires the vertical component of velocity also. It can be shown that path taken by this body is parabola using the equations of motion of kinematics. We can write the equations for the displacement along x-axis and y-axis and further substituting the value of the time from the x-axis equation in the y-axis equation, it can be proved that its path is parabola.

At any instant of the path, velocity of the projectile is tangential to its path. It has both horizontal and vertical components and the angle made by the velocity vector with the horizontal can be calculated as shown below. We can also calculate the effective velocity of the projectile at that instant.


Application

Let us consider a body projected horizontally from the top of a tower. The line joining the point of projection and the striking point of ground makes an angle of 45° with the horizontal.  What is the displacement of the body?

Solution

As the angle made at the horizontally is 45 degree, according to trigonometry, its horizontal and vertical distances are equal. Being the displacement is a vector quantity with can calculate its resultant value using the parallelogram law of the vectors as shown below.


Application

Let us consider a ball is thrown horizontally from a staircase as shown with a initial velocity. We shall calculate how many steps it travels before it strikes the ground?

Solution

The ball is having only horizontal component of velocity and has no initial vertical component of velocity.

We can use equations of motion to find the displacement along x-axis and y-axis. There is no acceleration due to gravity impact on the x-axis and there is gravity acting along the y-axis. The total horizontal distance travelled by the body is equal to the multiplication of the number of the steps and the breadth of each step. Similarly the total vertical distance travelled by the body is equal to the multiplication of number of steps with the height of each step.

By substituting this values in the above equation is can solve the problem as shown below.


Application:

Two bodies are thrown horizontally with a two different initial velocities in mutually opposite directions from the same height. What is the time after which velocity is of the two bodies are perpendicular to each other ?

Solution:

Velocity is a physical quantity that is having both magnitude and direction and it has to be treated like a vector quantity. When two vectors are perpendicular to each other their scalar product becomes zero. Taking this into consideration and by writing the equations of velocity is for the two bodies after a specific time and equating their product to 0 with can solve the problem as shown below.


Problem

An object is thrown horizontally from a point it hits the ground at some another point. The line of sight from these two points makes an angle 60° with the horizontal. If acceleration due to gravity is 10 m/s Squire and time of flight is known what is the velocity of projection?

Solution

by writing the equations were displacement along x-axis and y-axis and by using a trigonometrical definition we can solve the problem as shown below.


Problem

Two paper screens are separated by a distance of hundred meter. A bullet fired through them. The hole made in the second screen is 10 cm below the hole made in the first screen. What is its initial velocity before it strikes the first screen?

Solution

Initially the bullet is having only horizontal velocity and its vertical component of velocity equal to 0. There is no impact of acceleration due to gravity along the x-axis and it is acting along the y-axis. Taking this things into consideration we can equate the distance between the two poles along the vertical direction is equal to the vertical distance travelled by the body during the journey time. We can also calculate the time taken by the body for this to happen using the equation of the displacement along the x-axis. By solving these equations together we can calculate the initial velocity of the projection as shown below.


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Velocity of Projectile and Problems on Projectile Motion

A projectile is a body that is having a two-dimensional motion. A body thrown, making an angle with the horizontal other than 90° can have a two-dimensional motion. We shall project the body with a initial velocity for this to happen. This initial velocity can be resolved into horizontal and vertical components. There is no influence of acceleration due to gravity along the horizontal direction and hence velocity component along the horizontal direction always remains constant. As the gravity acts along the Y direction, the velocity component along the Y direction keeps changing with respect to time. Here we are going to find out the effect to velocity of the projectile after a specified time.

Let us consider a body is projected with the initial velocity making an angle with the horizontal. We can calculate the horizontal and vertical components of velocities of the body using the equation of motion. Being these two are perpendicular components to each other, using the parallelogram law that starts with can find the effective velocity of the projectile as shown below. It can be noticed that at the maximum height vertical component of the velocity zero and hence the velocity of the projectile is equal to the horizontal component of velocity itself. And hence projectile will have least possible allows state the maximum height but that is not equal to 0. We can also calculate the change in the moment of the projectile between the two points of its initial and final journey.



Let us calculate the time after which initial and final the last is of the projectile are perpendicular to each other. We can write the velocity of the projectile in vector format as shown below. Both initial and final velocities of the projectile were returned the vector formats. We know that if two vectors are perpendicular to each other their scalar product has to be zero. Hence we can find the scalar product of the initial and final at the arts of the projectile and we can equate to 0. By doing this we can identify the time after which initial and final velocity of the projectile are perpendicular to each other as shown below.


Using the similar sort of concept we can also calculate the velocity of the projectile at half of its maximum height. Even in this case the horizontal component of velocity remains constant. We can find out the vertical component of the velocity at the half of the maximum height using the third equation of motion of kinematics as shown below. Anyway these two components are always perpendicular to each other and by using the parallelogram law sectors we can find the resultant vector as shown below.


Problem on velocity of projectile

A body is projected with a initial velocity and by making an angle with the horizontal. The body makes 30° angle with the horizontal after two seconds and then after one second it reaches the maximum height. What is the angle of projection and what is the speed of the projectile?

Solution

We know that horizontal and vertical components of elasticity different and we can find the angle between them using a trigonometrical  equation. By equating this value with the 30° with can get one relation. When the body reaches the maximum height the angle made by the body with the horizontal is zero and again by equating with this value with can get one more a question. By solving these two equations we can get the velocity of the projectile and angle of the projectile as shown below.


Problem on projectile motion

A football is kicked with a initial velocity of 19.6 m/s with an angle of projection of 45 degree by one player. On the goal line 67.4 m away from this player and from the direction of the click, another player start running to meet the ball at the same instant. What must be his speed to catch the ball before it lands in the goal post ?

Solution 

We know that as the ball is projected with a 45° angle it’s angle of projection helps the body to reach its maximum range and we can calculate the maximum range using its formula as the 39.2 m. And we can also calculate the time taken by the body to reach the ground using the time of fight formula as shown below. If the other player has to catch the ball he has to reach the ball before it strikes the ground in the same time that is equal to time of flight. As he is running on the ground with a uniform velocity we can use a simple formula that the distance that it has to cover is equal to product of velocity and time. Hence we can solve the problem as shown below.


Expression of range and maximum height in terms of constants of the parabolic equation

we can express range and maximum height in terms of constants of the parabola as shown below.




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Projectile Motion Range,Time of Flight and Maximum Height Equations

A body is said to be in translatory motion if all the particles of the body are having a similar kind of displacement and velocities. If this body is moving only along one direction, it is called one-dimensional motion. If the body is thrown with an angle, then it will have motion both along the x-axis and y-axis simultaneously and this kind of the motion is called two-dimensional motion. It is also called projectile motion.

If the body projected with an angle other than 90° with the horizontal then, it is called projectile. We can study displacement and velocity of this kind of motion simultaneously on both X and Y axis. We can also find out what is the maximum horizontal distance that the body can travel. This maximum horizontal distance that a projectile travels is called the range. The time taken by the projectile to reach the ground from the point of projection is called Time of flight. The maximum vertical distance that it can reach is called maximum height. Here we are going to derive the equations for all this values.We are ignoring all the impact of air friction on the motion of the body during this study for simplicity.

Let us consider that you are throwing a stone into the air but not vertically up. On this stone now there are two forces acting simultaneously. One force is the force that you have applied and another one is the gravitational force which is always acting in the vertically downward direction. The resultant motion of the body is due to these two forces. The gravitational force acting on the body is constant and because of its influence it finally comes down and then reaches the ground. Until the body reaches the maximum height, it is moving against the gravity and hence it’s velocity keep on decreasing. From the maximum height point on words the body is coming in the downward direction and gravity is also pulling it in the downward direction. Hence it’s velocity keep on increasing further.

Let us consider a body having a mass  stone with an angle to the horizontal with a initial velocity. The velocity of the body is in between X and Y axis and hence it can be resolved into components. The horizontal component of the velocity at along the x-axis and the vertical velocity is along y-axis. As there is no gravitational effect along the x-axis, this horizontal component of velocity always remains constant. But the velocity along the y-axis keep on changing with respect to time as gravity influences also changes.

Using the equations of motion we can write displacement of the particle along both x-axis and y-axis and by substituting the value of the time from the x-axis equation on the y-axis equation, we can prove that part of this body is projectile. It is as shown below.


The final equation of the displacement along y-axis represents a parabola according to mathematics. It is probably learned in the school level that the equation says that the body is having a simultaneous motion along both x-axis and y-axis and it takes a parabolic path.



We can further derive the equation for a time of flight. At the end of the time of the flight as the body is coming back to the ground, its displacement along y-axis is equal to 0. By equating the equation of the displacement to 0 with can get the equation for the time of flight as shown below. It can be also further noted that this time of flight is the sum of time of ascent and the time of dissent. Time of ascent is the time taken for the body to reach the maximum height and time of dissent is the further time taken by it to reach the ground.

Once if you know the equation of the time of flight, by putting dirty equation value in the displacement of the body along x-axis, we can calculate the total distance travelled by the body along x-axis. This total distance travelled by the body on the horizontal axis is called horizontal range.

It can be further notice that horizontal range will be maximum if the angle of projection is 45° . If we have noticed any of the athlete throwing a discus throw ball, he preferred to throw the ball by making an angle of 45° with the horizontal so that it can go for the maximum horizontal distance and hence he will be the winner.

It can also be proved that range is a projectile is equal for two angles of projection.


The horizontal component of the velocity always remains constant, as there is no gravitational impact in the direction. The vertical component of the velocity keeps on decreasing and by the time the body reaches the maximum height it will become zero. Hence the projectile at the maximum high it is having only horizontal component of velocity and it is the least possible velocity that the projectile can have at any point of the journey. Taking this into consideration and by substituting the value any question of motion we can derive the equation for the maximum height attained by the projectile as shown below.



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