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Mr. Powers

  1. Oct 9, 2012 #1
    A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 6.0m/s^2 . Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 12.0s , Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance.

    A) What is the maximum height above ground reached by the helicopter?

    B) Powers deploys a jet pack strapped on his back 5.0s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 1.0m/s^2. How far is Powers above the ground when the helicopter crashes into the ground?

    I got A as 700m already B is the one I need...
     
  2. jcsd
  3. Oct 9, 2012 #2
    How did you approach part B?
     
  4. Oct 9, 2012 #3
    I tried many ways I used Vf = Vi + at to try getting the time but that didn't work the time seemed to absurd and then I used Vf² = Vi² + 2ad to try getting distance but that was wrong to, kept getting negative numbers.
     
  5. Oct 9, 2012 #4
    How did you apply those equations, though, and what numbers did you use?
     
  6. Oct 10, 2012 #5
    Just use a few displacement functions.

    0 = 432 + 72th -.5(9.8)th2 //time for helicopter to reach the ground
    h = 432 - .5(9.8)52 //height at jetpack turn on
    h2 = h - 49t -.5t2 //Plug in th-5s and h


    ______________________________________
    If everyone is thinking alike, then someone isn't thinking

    Garrett Stauber​
     
    Last edited: Oct 10, 2012
  7. Oct 10, 2012 #6
    I still don't have B)
     
  8. Oct 10, 2012 #7
    The answer to b is h2. If you solve / insert your values the equations above it will give you the answer.
     
  9. Oct 10, 2012 #8
  10. Oct 10, 2012 #9
    I got approx. 239.5 m. There was a mistake in my 2nd equation, I assumed Powers began falling at G as soon as he stepped out, however his initial velocity allows him to continue upward for a while. So you have to calculate where he will be in 5s after stepping out (using a displacement function) and his velocity after those 5s (using v = v0 + at) Hint: His V0 is 72 m/s.
    .
     
  11. Oct 10, 2012 #10
    My number seems to high about 439m...
     
  12. Oct 11, 2012 #11
    I'll go through step by step then from the beginning for you.

    What to find:
    H1=Height at which the engine shut off
    V1=Velocity at H1
    H2=Maximum Height
    T=Time for Helicopter to reach the ground
    V2=Powers' velocity after 5s
    H3=Powers' Height after 5 s
    H4=Powers' Height at helicopter impact



    H1=0+0(12)+.5(6)(122)=432
    V1=0+(62)(12)=72
    H2=Max{432+72(t)-.5(9.8)t2} = 696.5
    T→ 0=432+72(t)-.5(9.8)t2 → T=19.23s
    V2= 72 - 9.8(5) = 23
    H3=432+72(5)-.5(9.8)(5)2 = 669.5
    H4=669.5-23(19.23-5)-.5(1)(19.23-5)2=240
     
  13. Oct 11, 2012 #12
    It says 240m is wrong but I see how you got it.

    Btw thanks a lot for taking time to help me, much is appreciated.

    Not sure why 240m is wrong?
     
    Last edited: Oct 11, 2012
  14. Oct 11, 2012 #13
    I believe there is an incorrect negative sign in my last equation. See if you can find it, and try 895.5 for the answer.
     
    Last edited: Oct 11, 2012
  15. Oct 11, 2012 #14
    ∏assignment is due in 2 hours :S ,not many tries left either, 895.5? That's higher then the max height :P, but is the negative in h4 when its -.5 should be +?
     
  16. Oct 11, 2012 #15
    That is higher than the helicopters max height, but remember his acceleration changes to -1 after 5 seconds while the helicopter acceleration stays at -9.8. Powers Max Height is actually 934, he is still climbing when the chopper hits the ground. The (-) initial velocity was incorrect.
     
  17. Oct 11, 2012 #16
    I love you so much man, 900m was correct, thank you sooooooo much!! The explanations were wonderful...can't thank you enough!
    THANKKKKKK YOU
    THANK YOU
    THANKKKKKKK YOU
    THANK YOU!!!!@$!@$!@$!@
     
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