# Trajectory Motion Using Calculus

1. Apr 29, 2007

### cybershark5886

At this point in my project I'm merely trying to work out the correct formula to put together to find the x and y components of a trajectory travelling in a parabollic path.

My teacher showed me two different forms and said for me to try to figure out which one was best for my specific application, before I apply it. One form took an integral approach and had the components as such:
[Delta]Y = Vyit + 1/2at^2
[Delta]X = Vxit + (1/2at^2) [since no acceleration in x direction only Vxit is relevant]

The other took a derivative approach using the basic parabolic formula ax^2+bx+c=0 and takes this form:

s = at^2 + vt + c = 0

With the x and y components being:

x) s = vt
y) s = at^2 + vt

As you can see the last form is much simpler and straight forward. The thing is though I need to measure initial velocity. Also my teacher didn't know which form was more suitable for my application. I tried getting her to explain the difference between the first approach using 1/2at^2 and the second aproach (on the y component) using only at^2. What happened to the 1/2 in front? She said something about the first being a result of the integral approach of integrating at which would then produce 1/2at^2. That makes sense but why does the formula s = at^2 + vt not havethe 1/2 in front?

One last question concerning this she wrote down two distance formulas on my paper while we were discussing this but she was going too fast that I didn't quite catch the difference.

She wrote d = 1/2at^2 on the left and then right beside it d=16t^2. I think she said the 16t^2 part had something to do with taking gravity into account. Can any one clear this up for me a bit? I would appreciate it.

Thanks,

~Josh

Last edited: Apr 29, 2007
2. Apr 29, 2007

### cybershark5886

Oops. I apologize, I just realized this probably is in the wrong forum section. I didn't see the sticky note at first. Sorry never posted here before. Could a moderator move this to the proper forum?

3. Apr 29, 2007

### robphy

s = at^2 + vt + c is physically incorrect.

To see what the coefficient of t^2 represents, take two derivatives of the displacement s, written as s=At^2+Bt+C, with respect to t.

4. Apr 30, 2007

### cybershark5886

To see what the coefficient of t^2 represents, take two derivatives of the displacement s, written as s=At^2+Bt+C, with respect to t.

Ok, the first derivative would be 2At + B and the second derivative would be 2A. I'm not quite sure of the consequences of that, but would you mind helping me fix the formula if it is wrong?

5. Apr 30, 2007

### robphy

So, the "second derivative of displacement s" (the "acceleration a") is equal to 2A. So, what is the coefficient A in s=At^2+Bt+C?

6. Apr 30, 2007

### cybershark5886

So, the "second derivative of displacement s" (the "acceleration a") is equal to 2A. So, what is the coefficient A in s=At^2+Bt+C?

It would be 1. But where are you going with this?

Let me just throw some dummy values out there.

Say my initial velocity is 15 meters/second and the duration of time in the air is 10 seconds.

Using the y position formula [Delta]Y = Vyit + 1/2at^2

Then it should yield: [Delta]Y = 15*30 + 1/2(-9.81)(10)^2
Which reduces to: 450 - 490.5= - 40.5 meters

Now how can I find its highest point in the trajectory though? Is that formula telling me that the projectile lands 40.5 meters below the firing height or that it went up to a peak of 40.5 meters and came back down?

Last edited: Apr 30, 2007
7. Apr 30, 2007

### robphy

No it would not be.
(The coefficient A (the "coefficient of t^2") is that thing that multiplies t^2 in that term.)

Since a=2A, then the coefficient A=(1/2)a... which explains that factor of (1/2) that you asked about in the beginning.