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Mathematics 14 Online
OpenStudy (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 .

OpenStudy (anonymous):

*58.8 m/s^2

OpenStudy (anonymous):

i found the height when the fuel is exhausted which is = to 1880

OpenStudy (anonymous):

the velocity at that point i think is 470 m/s

OpenStudy (anonymous):

i think i have to find the distance it went after that and add it?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

that 1880, how did you come to that

OpenStudy (anonymous):

1881.6 lol

OpenStudy (anonymous):

yea rounded wrong LOL

OpenStudy (anonymous):

470.4 m/s correct

OpenStudy (anonymous):

I must complain about this question - it doesn't take into account the loss of mass as the fuel is ejected.

OpenStudy (anonymous):

yea physics is a pain.. LOL

OpenStudy (anonymous):

hey man it's simple physics lol. if you take mass into account, school students are doomed

OpenStudy (anonymous):

i was using this formula Xf=Xi+(Vx)i Dt + 1/2Ax(Dt)^2

OpenStudy (anonymous):

when ever i see a physics problems i get anxiety.. =(

OpenStudy (anonymous):

ttt luv the pic

OpenStudy (anonymous):

anyway, what's the distance traveled after fuel ran out?

OpenStudy (anonymous):

i couldnt figure it out

OpenStudy (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?

OpenStudy (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

OpenStudy (anonymous):

a=-9.8m/s^2 negative since acceleration downwards. gravity

OpenStudy (anonymous):

uh huh!

OpenStudy (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

OpenStudy (anonymous):

what do i plug in for S?

OpenStudy (anonymous):

that's what you need to find. plug everything in to find S

OpenStudy (anonymous):

0=470.6^2 + 2 (-9.8)S

OpenStudy (anonymous):

yeap

OpenStudy (anonymous):

11299

OpenStudy (anonymous):

yea, looks reasonable add this S to the S before(1881.6) you get max height

OpenStudy (anonymous):

damn man youre a genius!

OpenStudy (anonymous):

it took me an hour to figure out the first part =(

OpenStudy (anonymous):

don t worry you will get used to it very soon, just keep practicing. All those problems more or less are the same

OpenStudy (anonymous):

thanks! when can i buy you lunch!? LOL

OpenStudy (anonymous):

lol what do you think, I eat physics for lunch hahaha. just kidding. I'm off to bed now. see ya.

OpenStudy (anonymous):

haha thanks good night!

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