Lukey_R
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A rocket is launched vertically from the ground at 53.9 ms^-1.


Find the length of time for which the rocket is more than 88.2m above the ground.

I keep getting 49/11 s but the answers 7s. Anyone know how to get 7s?
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the bear
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yes... you need to use SUVAT
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Fractite
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yeah use suvat... consider u=0 and s=88.2

also, is it the length of time that the rocket is above 88.2m? as in, it goes up then down?
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Lukey_R
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(Original post by the bear)
yes... you need to use SUVAT
I am, hence the 49/11 s. could you tell me how?
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Lukey_R
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(Original post by Fractite)
yeah use suvat... consider u=0 and s=88.2

also, is it the length of time that the rocket is above 88.2m? as in, it goes up then down?
Thats what i assumed, i just copied the question straight from the book
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the bear
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h = 53.9t -4.9t2

put in a suitable value for h and solve quadratically. the difference between your two values of t should be 7 seconds.
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DFranklin
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#7
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(Original post by Lukey_R)
I am, hence the 49/11 s. could you tell me how?
You should know how to form an equation of the form at^2+bt+c for the height of the rocket above the ground. Then find the 2 roots of at^2+bt+c = 88.2, inbetween these roots will be the time it is above 88.2
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timif2
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No it’s more simple than you think

U = initial velocity = takeoff velocity of the system (rocket) = 53.9

You’re given everything except the final velocity. Use suvat
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DFranklin
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(Original post by erinls2)
~snip~
Please don't post full solutions. (See the Posting Guidelines for more details on how you should answer questions).
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erinls2
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(Original post by DFranklin)
Please don't post full solutions. (See the Posting Guidelines for more details on how you should answer questions).
Ah, I'm sorry, I wasn't aware of this - I have deleted it Thanks for pointing that out!
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Lukey_R
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#11
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Yes, it does. Thank you!
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