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cybershark5886
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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
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
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