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i found the height when the fuel is exhausted which is = to 1880
the velocity at that point i think is 470 m/s
i think i have to find the distance it went after that and add it?
that 1880, how did you come to that
yea rounded wrong LOL
470.4 m/s correct
I must complain about this question - it doesn't take into account the loss of mass as the fuel is ejected.
yea physics is a pain.. LOL
hey man it's simple physics lol. if you take mass into account, school students are doomed
i was using this formula Xf=Xi+(Vx)i Dt + 1/2Ax(Dt)^2
when ever i see a physics problems i get anxiety.. =(
ttt luv the pic
anyway, what's the distance traveled after fuel ran out?
i couldnt figure it out
im thinking i have to use this formula V^2 = Vo^2+2a(y-yo) Vo is the initial velocity which is 470 m/s?
ok, take upwards as possitive direction at the point where feul ran out, u=470.6m/s, a=-9.8m/s^2m, v=0m/s at max point. use v^2=u^2+2aS
a=-9.8m/s^2 negative since acceleration downwards. gravity
so plug them in the equation, you'll find distance from no more gas point to max point, add this to distance from ground to ran out point to get max height
what do i plug in for S?
that's what you need to find. plug everything in to find S
0=470.6^2 + 2 (-9.8)S
yea, looks reasonable add this S to the S before(1881.6) you get max height
damn man youre a genius!
it took me an hour to figure out the first part =(
don t worry you will get used to it very soon, just keep practicing. All those problems more or less are the same
thanks! when can i buy you lunch!? LOL
lol what do you think, I eat physics for lunch hahaha. just kidding. I'm off to bed now. see ya.
haha thanks good night!