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Gravity on another planet

  1. Jan 17, 2016 #1
    1. The problem statement, all variables and given/known data
    You are touring a distant planet on which the magnitude of the free-fall acceleration is 65.0 % of what it is on Earth. For a little excitement, you jump off a precipice 500 m above the planet's surface. After 5.00 s of free fall, you ignite the jet-pack on your back, changing your acceleration to some new, constant value for the remainder of the fall. You reach the ground 26.0 s after igniting the jet-pack.
    At what speed do you hit the ground?
    2. Relevant equations
    d=vt+1/2at^2
    v=vi+at

    3. The attempt at a solution
    oykPOpI.jpg

    Can someone tell me what I'm doing wrong? I've entered this onto my homework website, but it's telling me my answer is wrong, or perhaps I've might've missed something. I've retried this question multiple times, first time getting 0.441 m/s, second time entering 0.449 m/s. I'm on my last few tries, and every time I enter a wrong answer I lose marks.
     
  2. jcsd
  3. Jan 17, 2016 #2
    i'm pretty sure your thought process is correct, the problem here is probably significant figures, rounding mistakes or using "wrong" values for constants (g)...
    first check the problem to see if it states ammount of significant figures you should use, then check if you are using your teachers preferred value for g (some use 9.8 other 9.81 other 10)
    lastly dont punch in any values untill the very end, make sure you use the calculator's answer recall for every operation, minimizing any number rounding induced mistake
    in the end replace the value(s) you got (into yoir equations) to see if it is consistent with the problem.
     
  4. Jan 17, 2016 #3
    I followed all that, answer is still wrong, as it comes to 0.448 m/s even without rounding numbers at all throughout the problem. The velocity should be positive, correct?
     
  5. Jan 18, 2016 #4

    ehild

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    The speed was asked, it can not be negative.
    0.448 m/s should be correct.
     
  6. Jan 18, 2016 #5
    it can be negative (if velocity was asked). if you take the downward y direction as negative, he would arrive with negative velocity. maybe that is the problem here, your teacher assumed you would take gravity acceleration as negative since its downwards, and the new jetpack acceleration as positive since its upwards.
    if that is to be the case I would take it up with the teacher to ensure you dont lose any marks because there is absolutely nothing wrong in choosing downward direction as postive
    it would be a big mistake on your teacher's side to ask for speed wanting velocity.

    anyway, have you tried reversing the steps? i mean, using the values you got does it trace back to 500m height cliff?
     
    Last edited: Jan 18, 2016
  7. Jan 18, 2016 #6

    SammyS

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    The velocity should be positive, to be consistent with your overall choice of signs. (Downward velocity, acceleration and displacement were all positive in your work.)

    The resulting velocity at impact is quite sensitive to the value you use for g.

    A value of g = 9.8 m/s2 gives a velocity of 0.487 m/s at ground level.

    The very crude value of g = 10 m/s2 gives a velocity of -0.288 m/s at ground level. That's an upward velocity which at first puzzled me. It corresponds to having a sufficiently deep hole for you to fall into, rather than impacting the ground. You will descend into the hole and then rise to ground level 26 seconds after deploying the jet-pack. With no hole, you would impact the ground in less than 26 seconds.
     
    Last edited: Jan 25, 2016
  8. Jan 24, 2016 #7
    Oh, I used 9.81 as my gravity, perhaps that may have effected my answer. Though I still haven't checked with my prof. Thanks for your input though.
     
  9. Jan 25, 2016 #8

    mfb

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    I get 0.4478 with g=9.81 m/s^2 (WolframAlpha query), within three significant figures that agrees with the two values posted here.

    With g=10 m/s^2 the solution is a negative velocity which does not make sense - you have to go below ground level within those 26 seconds - so clearly you should not use 10.
     
  10. Jan 25, 2016 #9
    Alright so my teacher gave us the solution to this problem, and instead of using 9.81 m/s^2, she first solved the problem with symbols and afterwards she plugged everything in while using g=9.8 m/s^2 instead of 9.81 m/s^2, which made a quite a difference in numbers. Though I will check with her if using 9.81 is fine cause it should be and the data given is all in 3 sig figs, so I decided to use 9.81 since that has 3 sig figs in which I could get my answer to have 3 sig figs. So in which her final answer came to be 0.486 m/s^2.
     
  11. Jan 25, 2016 #10

    ehild

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