Question

Modelling projectile motion with calculus.

Answer 1

This is a more general question, but today I learned how to model projectile motion of a simple object (without air resistance)

Answer 2

I learned how to express it parametrically, as well as show that it is a parabola.

Answer 3

i.e., I learned pretty much everything here
http://en.wikipedia.org/wiki/Projectile_motion

Answer 4

I have a few questions about projectile motion, however. If we wanted to model the projectile motion of a rocket assuming it had no air resistance and constant gravity, but it burned fuel that directly effected the magnitude of the vector, what would we do?

Answer 5

http://en.wikipedia.org/wiki/File:Ferde_hajitas1.svg
To clarify: Usually we assume that we fire the projectile at a initial velocity, and only the V(subscript y) is affected, due to gravity. However, we might have a rocket that increases force with a function of time.

Answer 6

say, if fuel burned at a rate u(x)=sqrt(t)*C, and this rocket burned fuel the whole time, how would we model the resulting projectile motion?

Answer 7

instead of having just an initial constant vector representing velocity, we would have to account for the addition of the fuel burning. how would we do so, however?

Answer 8

if you know the acceleration, we can work our way up towards a position function by integrating the acceleration twice

Answer 9

I have the resulting parametric equations with me, by the way.
x=V cos(a)t, for a projectile launched initial velocity V with angle a
y=(-1/2)gt^2+V sin(a)t

Answer 10

@amistre64 , do we just add on (or subtract) this force to the current ones?

Answer 11

the parametric equations do very little if we dont know how to get them to begin with, in my opinion

Answer 12

ok, let me type out the inital equations, lol.

Answer 13

This will take a while, lol

Answer 14

\(V_{x}(0)=\frac{dx}{dt}|_{t=0}=V_{0}∗\cos(\alpha)\)
\(V_{y}(0)=\frac{dy}{dt}|_{t=0}=V_{0}∗\sin(\alpha)\)

Answer 15

Gravity:
For x
\(F_{x}=0\)
\(0=m*\frac{d^2x}{dt^2}\)
For y
\(F_{y}=-mg\)
\(-mg=m*\frac{d^2y}{dt^2}\)

Answer 16

Solving for x:
\(x=c_{1}t+c_{2}\)
Solving for y:
\(y=-\frac{1}{2}gt^2+C_{3}t+C_{4}\)

Answer 17

Substitution of appropriate constants lets us get to our result, the parametric equations. This I completely (or I think I do) understand. What happens if we add a force \[u(x)=C_{fuel}*\sqrt{t}\], where u(x) itself is a force?

Answer 18

How do we know which components to add to \(F_{x}\), and \(F_{y}\)?

Answer 19

@amistre64 ?

Answer 20

Assuming that this force u(x) is applied on to the projectile from t=0

Answer 21

Also, the other initial conditions:
\(t=0,x=0,y=0,V_{0}\)

Answer 22

i believe you would decompose the Force vector and add the appropriate y to Fy and x to Fx

Answer 23

other than that, im not sure what that "physics" applications would allow :)

Answer 24

Any idea how I would decompose the force vector of u(x)?

Answer 25

And, any idea at which rate rocket fuel burns?

Answer 26

the force vector of u(x) should amount to the slope of the tangent at any given x

Answer 27

im not a rocket scientist lol

Answer 28

lol alright, well, have to go to bed, thanks for replying; I'll look into it tomorrow.

Answer 29

good lcuk

Answer 30

After a little thinking, I have decided to use a constant model for acceleration provided by fuel. The result is we now have the equations
\(F_{fuel x}=\cos(\arctan\frac{dy}{dx})*C_{fuel}\)
\(F_{fuel y}=\sin(\arctan\frac{dy}{dx})*C_{fuel}\)
Where \(F_{fuel x}\) is the force the fuel applies horizontally; \(F_{fuel y}\) is the force the fuel applies vertically, and \(C_{fuel}\) is the constant force that fuel applies. Now, @amistre64 , doesn't this come down to a differential equation?

Answer 31

@inkyvoyd , @amistre64 will be back soon.
Dont worry.

Answer 32

help with what?

Answer 33

mehh calculus in physics me no like @_@ i still gotta study for the REAL calculus though... sorry :/

Answer 34

seriously dude -_- http://openstudy.com/updates/4fa9dcbce4b059b524f5c516

Answer 35

i mean http://openstudy.com/study#/updates/4fa9dcbce4b059b524f5c516

Answer 36

\(F_{fuelx}=\cos(\arctan\frac{dy}{dx})∗C_{fuel}\)
\(F_{fuely}=\sin(\arctan\frac{dy}{dx})∗C_{fuel}\)
Initial conditions: \(t=0,x=0,y=0,V_{0}\)
resulting differential equations
\(\frac{d^2x}{dt^2}=\cos(\arctan\frac{dy}{dx})∗C_{fuel}\)
\(\frac{d^2y}{dt^2}=\cos(\arctan\frac{dy}{dx})∗C_{fuel}\)