# How to calculate projectile cordinates?

1. Jun 18, 2008

### leozeo

how to calculate projectile cordinates??

I have a ball of mass 1kg and 100N force is applied on it to move ball in upward direction at an angle of 45. So how do i calculate the coordinates to show the trajectory of the ball.

no air resistance.
ball just moves in upward direction when force applied on it.
gravity=10 m/s^2

please guide me how to solve this??
which formulas to be used??

--
leozeo

2. Jun 18, 2008

### Ed Aboud

Hi leozeo, welcome to PF

First off, take the following equation of motion $$V = u + at$$, where u=initial velocity, v=final velocity, a=acceleration, t=time, for the speed at any given time. Take this equation for your x direction, i.e. its speed horizontally. And take it in your y direction i.e. its speed vertically. So you will have $$V_x = u_x + a_xt$$ and $$V_y = u_y + a_yt$$.
Now try this for its displacement using the formula $$s = u t + \frac{1}{2} at^2$$.

Last edited: Jun 18, 2008
3. Jun 18, 2008

### leozeo

but i didn't get u

try to explain with any example

4. Jun 18, 2008

### ddelaiarro

My initial reaction is that you're missing some information. How long is the 100N force added to the ball?

What I'm getting at is this:

- If the 100N force is constant, then the ball will travel in 45* trajectory at a certain rate of acceleration (do a simple FBD and you'll see this phenomenon).

- Otherwise, you need to know how long the 100N force is being imparted on the 1kg mass to figure out a velocity. That velocity will then be effected in the y-direction by gravity (velocity will begin to decrease, hit zero, and then become negative) once the force is removed.

Is this an impact force (how long is the impact) or a constant force?

5. Jun 18, 2008

### leozeo

it just acts once on the ball. and consider that time as 0.

6. Jun 18, 2008

### Mech_Engineer

Then the ball never moves, it stays sitting on the ground because no acceleration is applied to it.

7. Jun 18, 2008

### ddelaiarro

Beat me to it. Force (resulting in a mass's acceleration) has to be imparted on the object for some period of time, no matter how small, for displacement to occur. See Newton's First Law.

8. Jun 18, 2008

### leozeo

as the ball thrown upwards with some force the ball remains in motion until it reaches its maximum height.

after that point it starts coming down.

9. Jun 18, 2008

### leozeo

instead of writing this much theory

10. Jun 18, 2008

### ddelaiarro

We're trying to do that. It's basically simple kinematics equations, but you need to determine initial velocities. And, to do that, you need a time that the force is acting on the ball.

11. Jun 18, 2008

### leozeo

what i have done is this

and now i want to replace that velocity with force and mass.

and based on that force and mass the ball should move

--
Leozeo

12. Jun 18, 2008

### Mech_Engineer

We've barely covered any theory at all, you want us to spoon-feed you an answer which isn't what this forum is about.

The equations you need are in the very first reply in this thread. You need to understand the application of the equations to properly utilize them, otherwise you'll never undestand what it is you're doing in the first place.

Given this equation: $$s = u t + \frac{1}{2} at^2$$

And this equation: $$F = m a$$

Perhaps you could try a simple substitution, like substituting $$\frac{F}{m}$$ for $$a$$ in the first equation?

Last edited: Jun 18, 2008
13. Jun 18, 2008

### leozeo

14. Jun 18, 2008

### ddelaiarro

The link you just provided show and angle and a VELOCITY, not a force. Given an initial velocity, as shown in the link, you're looking at a simple kinematics equation suing some of the equations already shown.

15. Jun 18, 2008

### leozeo

PHP:
and now i want to replace that velocity with force and mass.

and based on that force and mass the ball should move

--
leozeo

16. Jun 18, 2008

### ddelaiarro

I guess I'll write up a more thorough answer here to help you understand what we're getting at.

First off, you've got to remember you kinematic equations:

$$d=v_{i}+\frac{at^{2}}{2}$$

$$v^{2}_{f}=v^{2}_{i}+2ad$$

$$v_{f}=v_{i}+at$$

$$d=\frac{v_{i}+v_{f}}{2}t$$

In these equations,

$$d$$ = displacement
$$v_{i}$$ = initial velocity
$$v_{f}$$ = final velocity
$$a$$ = acceleration
$$t$$ = time

Also, remember that Force, F, is a function of mass, m, and acceleration.

$$F=ma$$

You've given us a Force and a mass. From this information, we can get an acceleration. That's two variables (mass and acceleration) that are defined. You'll notice that all of the equations above use four varibales. That means that, in order to solve for an unknown variable in a kinematic situation, you need at least three defined variables.

What we're saying is that you don't have enough information to calculate an answer.

17. Jun 18, 2008

### leozeo

thanks a ton

its of great help and i 'll try to implement these suggestions in my application

--
Leozeo

18. Jun 21, 2008

### Larbear

There is also another method of calculating range.
Where X=Vi$$^{}2$$Sin(2$$\Theta$$)/-g

X=Range
Vi=Initial Velocity
$$\Theta$$=Angle
G= Acceleration due to Gravity

However again without a method for calculating the Initial Velocity
you don't have quite enough information! If you could get flight time it would be possible to calculate.

19. Jun 23, 2008

### leozeo

if the object is placed at rest then the initial velocity will be 0

so consider initial velocity as 0.

--
leozeo

20. Jun 23, 2008

### ddelaiarro

leozeo -

You're still missing information. You need to know either the time period that the force is acting on the mass, the distance the mass travel (either x- or y-direction), or the velocity after the force has acted upon it.

Remember that acceleration is the change in velocity over the change in time

$$\frac{\partial V}{\partial t}$$

By this definition, acceleration cannot be 'instantaneous,' but must exist over some finite time element, not matter how short. In the definition above, the change in velocity would be due to the force acting on the mass.

I think you can see what I'm getting at in the equations I mentioned in this post.