anonymous
  • anonymous
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8 m/2^2. The acceleration period lasts for time 8.00 S, until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height Vmax reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 m/s^2 .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
*58.8 m/s^2
anonymous
  • anonymous
i found the height when the fuel is exhausted which is = to 1880
anonymous
  • anonymous
the velocity at that point i think is 470 m/s

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anonymous
  • anonymous
i think i have to find the distance it went after that and add it?
anonymous
  • anonymous
yep
anonymous
  • anonymous
that 1880, how did you come to that
anonymous
  • anonymous
1881.6 lol
anonymous
  • anonymous
yea rounded wrong LOL
anonymous
  • anonymous
470.4 m/s correct
anonymous
  • anonymous
I must complain about this question - it doesn't take into account the loss of mass as the fuel is ejected.
anonymous
  • anonymous
yea physics is a pain.. LOL
anonymous
  • anonymous
hey man it's simple physics lol. if you take mass into account, school students are doomed
anonymous
  • anonymous
i was using this formula Xf=Xi+(Vx)i Dt + 1/2Ax(Dt)^2
anonymous
  • anonymous
when ever i see a physics problems i get anxiety.. =(
anonymous
  • anonymous
ttt luv the pic
anonymous
  • anonymous
anyway, what's the distance traveled after fuel ran out?
anonymous
  • anonymous
i couldnt figure it out
anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
a=-9.8m/s^2 negative since acceleration downwards. gravity
anonymous
  • anonymous
uh huh!
anonymous
  • anonymous
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
anonymous
  • anonymous
what do i plug in for S?
anonymous
  • anonymous
that's what you need to find. plug everything in to find S
anonymous
  • anonymous
0=470.6^2 + 2 (-9.8)S
anonymous
  • anonymous
yeap
anonymous
  • anonymous
11299
anonymous
  • anonymous
yea, looks reasonable add this S to the S before(1881.6) you get max height
anonymous
  • anonymous
damn man youre a genius!
anonymous
  • anonymous
it took me an hour to figure out the first part =(
anonymous
  • anonymous
don t worry you will get used to it very soon, just keep practicing. All those problems more or less are the same
anonymous
  • anonymous
thanks! when can i buy you lunch!? LOL
anonymous
  • anonymous
lol what do you think, I eat physics for lunch hahaha. just kidding. I'm off to bed now. see ya.
anonymous
  • anonymous
haha thanks good night!

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