## Projectile motion derivation

In physicsassuming a flat Earth with a uniform gravity fieldand no air resistancea projectile launched with specific initial conditions will have a predictable range. The following applies for ranges which are small compared to the size of the Earth. For longer ranges see sub-orbital spaceflight.

The maximum horizontal distance traveled by the projectileneglecting air resistance, can be calculated as follows: . If y 0 is taken to be zero, meaning that the object is being launched on flat ground, the range of the projectile will simplify to:. Ideal projectile motion states that there is no air resistance and no change in gravitational acceleration.

This assumption simplifies the mathematics greatly, and is a close approximation of actual projectile motion in cases where the distances travelled are small.

Ideal projectile motion is also a good introduction to the topic before adding the complications of air resistance. A launch angle of 45 degrees displaces the projectile the farthest horizontally. This is due to the nature of right triangles. First we examine the case where y 0 is zero. The horizontal position of the projectile is. We are interested in the time when the projectile returns to the same height it originated.

Let t g be any time when the height of the projectile is equal to its initial value. The first solution corresponds to when the projectile is first launched. The second solution is the useful one for determining the range of the projectile. Plugging this value for t into the horizontal equation yields. Applying the trigonometric identity. Once again we solve for t in the case where the y position of the projectile is at zero since this is how we defined our starting height to begin with.

### Projectile motion – Derivation of Projectile motion equations for class 11

Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic manipulation. The square root must be a positive number, and since the velocity and the sine of the launch angle can also be assumed to be positive, the solution with the greater time will occur when the positive of the plus or minus sign is used. Thus, the solution is. In addition to air resistancewhich slows a projectile and reduces its range, many other factors also have to be accounted for when actual projectile motion is considered.

Generally speaking, a projectile with greater volume faces greater air resistancereducing the range of the projectile.As I throw a stone like this and it goes off my hand then I am no more applying any force on that object. But there will be a force omnipresent acting on that stone. That is Gravity or Gravitational force applied by the earth. And this force directs vertically downwards pointing to the center of the earth.

If gravity was not present that stone would continue to move in a straight line maintaining the same angle with the horizontal plane. But this downward pull of gravity takes it down.

On one hand, that stone tries to maintain a constant velocity along the horizontal as there is no force on it along that direction. On the other hand, it gets a downward pull caused by gravity and faces an acceleration equal to g acceleration due to gravity.

Under these 2 movements, it takes a specific path, which we will discuss now. When an object is in flight after being projected or thrown then that object is called a projectile and this motion is called Projectile Motion. Example, the motions of a cricket ball, baseball. The motion of a projected object in flight is known as projectile motion which is a result of 2 separate simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration as no force acting in this direction and the other along the vertical direction with constant acceleration due to the force of gravity.

In the next sections we will discuss and derive a couple of projectile motion equations. We will cover here Projectile Motion Derivation to derive a couple of equations or formula like:. Say an object is thrown with uniform velocity V 0 making an angle theta with the horizontal X axis. Air resistance is taken as negligible.

This is an equation representing parabola. So we can say that the motion path of a projectile is a parabola. So once thrown a projectile will follow a curved path named Parabola. When the projectile reaches the maximum height then the velocity component along Y-axis i.

Say the time required to reach this maximum height is t max.

Derivation of Time of Flight, Horizontal Range, Maximum Height of a Projectile

So this is the equation for the time required to reach the maximum height by the projectile. We know that when the projectile reaches the maximum height then the velocity component along Y-axis i.

Here we will use the horizontal component of the velocity only as the effective velocity to traverse this horizontal path. We will add some numerical as well very soon. Skip to content.

Derivation of Projectile motions equations or formula. Projectile motion — Derivation of Projectile motion equations for class By Anupam M.We have worked several examples of simple projectile motion — meaning that the acceleration of gravity was constant and vertical, and there was no acceleration in the horizontal direction.

In particular, there is no air resistance. I simply handed us four equations and used them in a first … second … and third post. I said I would show how to derive them. It was junior year in high school that I learned the equations for position and speed as a function of a constant acceleration. This is about as elementary as it gets in calculus, but when it was all new to me, it was a thrill to see what it could do for me in physics.

I will actually derive them in two slightly different ways. We want the acceleration to be constant, so we write a instead of a t :. There is yet another way to do this; use the fundamental theorem of integral calculus, which says:.

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It relates the integral of f to the area under the curve f. Yes, I have changed the variable of integration. For constant acceleration,and we get….

Drop in our formula for v t — excuse me, …. Finally, for simple projectile motion, we take the horizontal component of acceleration to be zero, and the vertical component of acceleration to be -g. So im a bit late to you post, a year and then somebut I came across this when preping for my final exam. I just wanted to drop a line and say you are very skilled in explaining these concepts in a systematic and clear way. It was a nice read and refresher. Neat explanation. The point here is that what the differential equation contains isand the expression for this must be written as a sum in xn.

For this reason we shift the dummy variable in the series for the first derivative by You are commenting using your WordPress. You are commenting using your Google account. You are commenting using your Twitter account. You are commenting using your Facebook account. Notify me of new comments via email. Notify me of new posts via email.

Here we go. First Derivation Acceleration is the derivative of velocity… and that means velocity is the integral of acceleration. We want the acceleration to be constant, so we write a instead of a t : and do the integral, supplying a constant of integration when the integral sign disappears:.

Next, velocity is the derivative of position… so position is the integral of velocity. That is, We plug in our formula for v t … We do the integral, getting another constant of integration…. For constant acceleration,and we get… which is… so finallyjust as before.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up. I'm currently learing kinematics, specifically projectile motion and as an example in my textbook is a bullet fired at some angle. Nick Basically you are confusing two different coordinate systems. In the coordinate system where you decomposed g into two components, there the question of x and y doesn't seem valid because you have transformed the cartesian coordinate system into a different one.

You would have new x and new y there and you have to stick with one while solving the kinematics. You are free to choose any of the coordinate system however analysis in the reference frame more technical term where you decomposed the gravity into two components is a bit more typical the reference frame itself is changing orientation and you would have to take that into account if you are inclined to solve in that frame.

A good and similar example of this would be to consider motion on an inclined plane.

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When a body is undergoing motion "along" the inclined plane, it is far more convenient to solve in a reference plane which has axis x and y parallel and perpendicular to the inclined plane. And thus since you fixed your reference frame now, you have to resolve all the forces etc. This is purely a simplification or a transformation from the normal cartesian system to the inclined plane one.

And as mentioned above, you can also have polar coordinates. Before solving any kinematics or dynamics problem, fix the coordinate system and this fixes your convention too - positive and negative and then start resolving forces, velocities etc. A natural question arises then how to choose which frame to solve in and avoid this confusion of mixing up two or more frames? The answer is practice and experience.

You would yourself start seeing that its tedious and highly complex to solve in one frame and thus would choose to go with the simplest solving. That's Physics, not mathematics. The acceleration due to gravity is downward, as you said. Now, we're in a 2D plane, so we can decompose this vector in two components. There are different ways of doing this. One can use a Cartesian coordinate system, i. Alternatively, we can use polar coordinates.

Note that, for this coordinate system, the decomposition depends on where you define your origin. This is not the case for a Cartesian coordinate system. You could draw a picture of this choosing, for instance, the origin at the position from where the stone is thrownthis might make it more clear why the decomposition depends on the location of the origin in polar, but not in Cartesian, coordinates.

Asked 5 years, 11 months ago. Active 4 years ago. Viewed 1k times. Nick Nick 2 2 bronze badges. Active Oldest Votes.

### Range of a projectile

Assess the effect of angle and velocity on the trajectory of the projectile; derive maximum height using displacement. Projectile motion is a form of motion where an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.

In a previous atom we discussed what the various components of an object in projectile motion are. In this atom we will discuss the basic equations that go along with them in the special case in which the projectile initial positions are null i. The time of flight of a projectile motion is the time from when the object is projected to the time it reaches the surface. In projectile motion, there is no acceleration in the horizontal direction.

The horizontal velocity remains constant, but the vertical velocity varies linearly, because the acceleration is constant. We can use the displacement equations in the x and y direction to obtain an equation for the parabolic form of a projectile motion:.

Using this we can rearrange the velocity equation to find the time it will take for the object to reach maximum height. From the displacement equation we can find the maximum height.

Using this we can rearrange the parabolic motion equation to find the range of the motion:. Range of Trajectory : The range of a trajectory is shown in this figure.

Projectiles at an Angle : This video gives a clear and simple explanation of how to solve a problem on Projectiles Launched at an Angle.

I try to go step by step through this difficult problem to layout how to solve it in a super clear way. Best wishes. Tune into my other videos for more help. In projectile motion, an object moves in parabolic path; the path the object follows is called its trajectory. We have previously discussed projectile motion and its key components and basic equations. Using that information, we can solve many problems involving projectile motion. Projectile motion is when an object moves in a bilaterally symmetrical, parabolic path.

Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity. Refer to for this example.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up. Do explain the reason. Please explain everything using calculus integration as much as possible and usage of equation of parabola would also help.

These equations count along any path. This is just how the throw was done, and how it flies doesn't alter how it started - these parameters will be different for other throws but do not change continuously during the flight. You might need to insert one equation into another and a lot of algebra can be needed in some cases. But you will almost always have enough equations to take from you have 8 to choose from. You might have to calculate this time first.

I am not sure this is a correct formula for the trajectory length the length of the flight path. But I would try to find the actual trajectory length by first making a trajectory equation see here for example ; go down to the "Parabolic Trajectory" headline and then to use the length formula on it see here. Those expressions are:. What you already have here does not look right to me, but I can't be sure since I don't remember this expression by heart without going through these steps to obtain it.

This relation can be broken down into two components :. Range for a projectile is defined as the horizontal distance covered by the projectile in that time period i.

## Projectile Motion Formula

The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Derivation of equations projectile motion Ask Question. Asked 2 years, 11 months ago.

Active 2 years, 2 months ago.

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Viewed 23k times. Mathejunior Mathejunior 1 1 gold badge 3 3 silver badges 10 10 bronze badges. It will be the same thing as throwing a projectile in deep space. It will follow a straight path throughout its motion.

Otherwise your post will be closed. Show your attempt and failures to derive the equations. But I cannot understand it well. Hence I posted it here. Active Oldest Votes.

Steeven Steeven Things will be more clear if we separately compare the displacements along the two axes. Range : Range for a projectile is defined as the horizontal distance covered by the projectile in that time period i. And in this time, the horizontal distance covered by the projectile will be its range.

We are interested only in the second case.

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Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up. Do explain the reason. Please explain everything using calculus integration as much as possible and usage of equation of parabola would also help.

These equations count along any path. This is just how the throw was done, and how it flies doesn't alter how it started - these parameters will be different for other throws but do not change continuously during the flight. You might need to insert one equation into another and a lot of algebra can be needed in some cases. But you will almost always have enough equations to take from you have 8 to choose from.

You might have to calculate this time first. I am not sure this is a correct formula for the trajectory length the length of the flight path. But I would try to find the actual trajectory length by first making a trajectory equation see here for example ; go down to the "Parabolic Trajectory" headline and then to use the length formula on it see here. Those expressions are:.

What you already have here does not look right to me, but I can't be sure since I don't remember this expression by heart without going through these steps to obtain it. This relation can be broken down into two components :. Range for a projectile is defined as the horizontal distance covered by the projectile in that time period i. Sign up to join this community.

The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Derivation of equations projectile motion Ask Question. Asked 2 years, 10 months ago.

Active 2 years, 2 months ago. Viewed 23k times. Mathejunior Mathejunior 1 1 gold badge 3 3 silver badges 10 10 bronze badges. It will be the same thing as throwing a projectile in deep space.

It will follow a straight path throughout its motion.