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Kinematics, Throw-up problem

  1. Mar 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Marian, who is standing on her balcony, is surprised by a pigeon and throws a flowerpot up in the air at 2.1 m/s. It takes 3.0 s for the flowerpot to smash to the ground. The flowerpot experiences acceleration due to gravity of 9.81m/s2 [down].

    a) How high is Marian's balcony?
    b) How fast was the flowerpot moving just before it smashed to the ground?

    v1= 2.1 [up]
    a= -9.81m/s2 [up]
    t= 3.0 s

    2. Relevant equations

    Since acceleration is constant
    d = v1*t + .5*a*t*t
    v2[squared] = v1[squared] + 2*a*d

    3. The attempt at a solution

    First i used d = v1*t + .5*a*t*t
    d = 2.1*3.0 + .5*-9.81*3.0*3.0
    d = -37.8 m [up] or -38 m [up] ---> i round it to 2 sig figs here, right?

    Now here is where i get confused.

    This is negative displacement because the flower pot is flying most of it's way down to the ground. Right? What i am confused about is, is this the actual displacement from the balcony to the ground? Is Marian's balcony 38 m high?
    It just seems not right to me for some reason, it feels like i am missing some sort of step.

    If this is the answer, then for b) i use
    v2[squared] = v1[squared] + 2*a*d
    v2[squared] = 2.1[squared] + 2(-9.81)(-38)
    v2[squared] = 749.97[square root]
    v2 = 27.4 m/s or 27 m/s --->should i make it three or two figs. I think two figs, right?

    If everything is correct so far, why do i have positive velocity instead of negative. Should it not be negative?

    Thanks in advance and if i am totally wrong, please do correct me :)
    Last edited: Mar 9, 2011
  2. jcsd
  3. Mar 9, 2011 #2


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    welcome to pf!

    hi cidilon! welcome to pf! :smile:

    (have a square-root: √ and try using the X2 and X2 icons just above the Reply box :wink:)
    yes, Marian's balcony is 38 m high (plus the height of her hand when she throws) …

    the displacement is from the initial position :wink:
    if v2 = 27, v can be negative :wink:

    (and yes, 2 sig figs)
  4. Mar 9, 2011 #3
    Re: welcome to pf!

    So in this case it has to be negative right?
    Did i just not get the negative because for displacement i used -38m, instead of just 38 which should be used this time around?

    Thanks for the warm welcome :)
  5. Mar 9, 2011 #4


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    the question asked "how fast …", so the answer has to be positive :wink:

    (how you get to the answer, ie whether you measure up or down, is up to you … or down to you :biggrin: … but you must answer the question as asked!)
  6. Mar 9, 2011 #5
    Okay, so the answer is positive. Is my answer incorrect? Am i not answering the question asked? Sorry for my confusion!
  7. Mar 9, 2011 #6


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    √(27.4), to 2 sig figs, is correct :smile:
  8. Mar 9, 2011 #7
    I have to square root 27.4 again? I mean i got the answer by square rooting, why do i have to do it again? I have a feeling i started overcomplicating things again...
  9. Mar 9, 2011 #8


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    ohhh! i misread that …
    … as v2 ! :biggrin:
  10. Mar 9, 2011 #9
    Awesome, thanks! :)
  11. Mar 9, 2011 #10
    Sorry to post again but i am confused about something else as well.

    To get the second velocity i can also use v2 = v1 + a*t
    If i do it this way i get -27 [up] instead of 27 [up] as with previous equation.

    First, i am confused as to why i get different results? I understand that the question asked "how fast..." but doesn't -27 just mean that the motion is going in the opposite direction of the initial motion. So, the initial velocity was 2.1 m/s [up] and the second is 27 [down] or -27 [up] because it is going in the other direction. I am just confused that you said that when it says "how fast" it must be positive.

    Also if possible, please can you explain why i keep getting 27 with first equation and -27 with second. Which equation should i used? The only reason i think that might happen is that when i use this equation...

    v2[squared] = v1[squared] + 2*a*d
    v2[squared] = 2.1[squared] + 2(-9.81)(-38)

    ..for displacement i put -38 instead of 38, which probably should be there instead.
    Although i am not sure if i can set it as positive, if in my previous answer it was negative in relation to the balcony. So should it not remain negative here as well? Hmm..

    Please look this over, i want to be confident :)
  12. Mar 10, 2011 #11


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    hi cidilon! :smile:

    (just got up :zzz: …)

    if your equation has a square, then you can have either + or - …

    you have to look at the reality to see which one it is!

    this is english rather than maths …

    "how fast" is always positive :smile:

    (translated into maths, the question would ask "what is the magnitude of the velocity?")
  13. Mar 10, 2011 #12
    The only reason why i am concerned is that it just does not make sense that v2 is positive because it is flying [down] compared to v1 which is flying up. I understand that when i measure v2 it is accelerating down in the negative direction which makes it positive. Therefore my answer has to be positive... but my answer for v2 is it 27m/s [down]? I mean, that would make sense but as the answer was positive it can not really be [down] which is negative. If it is [up] then i am even more confused, as v1 is [up]. What do you mean by "if your equation has a square, then you can have either + or -"?


    I am also very sorry for bugging you so much, i hope i am not getting on your nerves!
  14. Mar 10, 2011 #13


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    you had …

    v2[squared] = 2.1[squared] + 2(-9.81)(-38)​

    … so v2 = ±√(2.1[squared] + 2(-9.81)(-38)) …

    both + and - are solutions to that equation :wink:
  15. Mar 10, 2011 #14
    I finally got it, thank-you so much!!
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