Trajectory Motion Using Calculus

[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 traveling 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 approach (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 anyone clear this up for me a bit? I would appreciate it.

Thanks,

~Josh

 

[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?

 

[robphy]

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

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.

 

[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?

 

[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?

 

[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?

 

[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?

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

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.