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Physics Problems -_-

PostPosted: Wed Oct 25, 2006 7:19 pm
by Sai
Hey all, I need some help. I am supposed to figure out how to find the max height a certain rocket reaches and i cant figure out the proper EQ. to use. Any help will be much appreaciated.

PostPosted: Wed Oct 25, 2006 7:41 pm
by Slater
ooh... are you supposed to take into account the change in mass that the rocket goes through?

PostPosted: Thu Oct 26, 2006 12:43 pm
by Sai
I think so.

PostPosted: Thu Oct 26, 2006 2:33 pm
by Puguni
Sai wrote:I think so.


I think it would be best if you actually copied the problem here.

PostPosted: Thu Nov 09, 2006 1:36 pm
by Sai
Sorry guys, i was busy and didnt get around to finding the stuff for the equation but we already launched the rocket so meh. Thanks for checking on it though.

PostPosted: Sat Nov 18, 2006 12:18 pm
by Dante
OOC: Don’t know if this helps, but rocket science is fun! First, in rocket science, F doesn’t equal m*a, you would have to use the more fundamental equation of F equaling the time derivative of momentum! YAY!!! Generally though, you could just be lazy and look it up like I did, Source: (The Cambridge Handbook of Physics Formulas by Graham Woan, Page 70 (Rocketry) Equation 3.94 the equation yields, (assuming your rocket starts out from rest)

[1]
vf =u*ln(Mi/Mf)


Where
vf= final velocity of your rocket when the engine dies
u=the velocity of the rocket exhaust (from the frame of reference of the rocket)
Mi=the initial mass of the rocket (not weight but it doesn’t matter because gravitational constant will cancel out)[with fuel]
Mf=the final mass of the rocket (without fuel)
ln is the natural log of what is in the paranthesis

Of course, this equation only gives the velocity of a rocket in space (no gravity). Generally the equation includes gravity, but we’re not looking for vf in a gravitational field, we’re only looking for hf. The purpose of this equation was to give us a vf for the standard equation,

[2]
KE=Mf*vf^2/2


As this is the kinetic energy of our rocket outside of a gravitational field, we can dump all that energy in at once into the particle, through the method of using energy to find height,

[3]
KE=PE; vf^2/2= g*hf


Where,
g=9.80665m/s^2=32 ft/sec^2

Solve for your hf in equation 3 after finding vf from equation 1 and you should be set for the final height of your rocket through use of the conservation of energy equation. Note that this equation assumes a constant gravitational field, which should work in your case, but for very high rockets you would have to modify your PE from Mf*g*hf to Mf*ME*G/hf. Where ME is the mass of the Earth, but you won’t have to worry about that, generally your final actual height will be less then the theory because of frictional effects which can be approximated to a certain extent but I doubt you will be required to consider friction for your rocket experiment, there is also a Corriolis effect and Centrifugal Effect depending on your longitude on the Earth but once again we’ll throw it out as negligible. Pretty sure everything is on target though,
Pascal