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Math Unit Problem Help Please

  1. Feb 6, 2005 #1
    Here is the problem:

    Homer knows that the rocket took 14 seconds to come down from its highest point. This means that the rocket was in the air for about 28 seconds before it hit the ground. Use the following equation from the movie to find out the initial velocity of the rocket in feet per second.

    s = 1/2at^2 + vt

    a= -32
    t= time in seconds
    s= altitude in feet
    v= initial velocity

    I belive (not too sure) the altitude will be 14x5280 (5280=1 mile in feet) but im not sure wether the time is 14 or 28 seconds. A little help please?
  2. jcsd
  3. Feb 6, 2005 #2


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    1.What movie??
    2.The TOTAL time has to be 28 seconds.
    3.How did u get that number...?

  4. Feb 6, 2005 #3
    The movie is October Sky
  5. Feb 6, 2005 #4
    so my full equation would be:

    14x5280 = 1/2(-32)(28)^2+v(28) ?

    Im still not sure what the altitude would be
  6. Feb 6, 2005 #5


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    How did u come up with 14 miles...??


    P.S.Did u know that
    [tex] v_{fin}=v_{init}+at [/tex]
  7. Feb 6, 2005 #6
    s is a function of time. at time = 14s, s is at its maximun value, however, you know nothing about the max_s coz the problem didn't mention about it..
    but if you put t=28, the rocket will hit the ground and you know the altitude is zero at that time, therefore, you better choose t=28
  8. Feb 6, 2005 #7
    Altitude = Highest point reached right? if that's the case the total time the rocket was in the air is 28, meaning 14 seconds going up, 14 coming down. If alititude does = highest point, then at 14 seconds the rocket hits its alititude, i just did 14x5280 as a guess i would say, any idea of how else to find the alititude?
  9. Feb 6, 2005 #8
    Heres the top half of the problem:

    Homer Hickim and his friends shoot off a rocket. A fire is reported 3 miles away from the launch pad (1mile = 5289 feet). Homer and his friends are accused of the fire.
  10. Feb 6, 2005 #9


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    Okay,now tell us HOW DID U GET THAT 5280 FEET...Are they given in the problem & and i don't see them...?


    EDIT:That changes everything...What didn't u say that in the beginning...??
  11. Feb 6, 2005 #10
    5280 feet = 1 mile
  12. Feb 6, 2005 #11
    Sorry for not mentioning it earlier,

    so would my correct equation to find the initial velocity be:

    14x5280 = 1/2(-32)(28)^2+v(28) ?
  13. Feb 7, 2005 #12
    i don't think this is a hw problem, therefore, I will give out the solution to cease the confusion
    [tex]\frac{1}{2}(-32)28^2 + 28v=0 [/tex]
    solve for v and you'll get:
    v=448 ft/s
    assuming no air drag...(which is unphysical under this speed)
  14. Feb 7, 2005 #13

    Doc Al

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    I believe that Demise messed up the statement of the problem in giving only the second half of the question. Here's the complete problem: https://www.physicsforums.com/showthread.php?t=62889

    What you've calculated is the vertical component of the initial velocity; but, as shown in post #8, it also has a horizontal speed.

    (Of course it's homework!)
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