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Interesting Physics Problem

  1. Feb 27, 2005 #1
    hey, i'm having a bit trouble solving this physics question! :mad:
    so if anyone can help me, here it is:
    A stone is thrown up past a window. The stone requires 0.20s to completely pass by the 1.4m tall window. How high above the window did the stone go? I'm aware that the average velocity is 7.0m/s using the formula Vave=d/t. However, from there...i'm clueless.
    Last edited: Feb 27, 2005
  2. jcsd
  3. Feb 27, 2005 #2
    You can use :
    [tex]v = v_0 -gt[/tex]
    [tex]y = y_0 + v_0t -g \frac{t^2}{2}[/tex]

    The v_0 and y_0 are the initial velocity and position and the latter is equal to zero in this case. You will need to determin the initial velocity v_0...this is easy because you kow that you have travelloed 1.4m in 0.2 seconds. Put this into the second equation and solve for v_0

    In order to determin the height above the window, just realize that the maximal height is acquired when the velocity is 0...

    What is this average velocity??? If you apply the normal definition [tex]v_{average} = \frac {v_{final} - v_{initial}}{2}[/tex] you will get ZERO because the inital and final velocity are equal to each other.
    Am i missing something ???

  4. Feb 27, 2005 #3
    Since i'm only in first level high school, i have not been aware of the formulas that were given. In terms of average velocity, i was given the formula that is was the division of distance over time, and therefore giving 7.0m/s of the stone passing the window. I am still unclear of what you are trying to say.

    ps) thank you for your assistance
  5. Feb 27, 2005 #4

    Doc Al

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    I believe jen333 was talking about the average velocity of the stone during its pass by the window (from bottom to top of window). [itex] V_{ave} = D/t = 7.0[/itex]m/s; true, but not that useful. Also, [itex] V_{ave} = (V_i + V_f)/2[/itex] for uniformly accelerated motion.
  6. Feb 27, 2005 #5


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    :rofl: Yes,use the definition for average velocity...
    [tex] \langle \vec{v}\rangle_{[t_{1},t_{2}]} =:\frac{1}{t_{2}-t_{1}} \int_{t_{1}}^{t_{2}} \vec{v}(t) \ dt [/tex]

    Choose [tex] t_{1}=0 ,t_{2}=t_{final} [/tex]

    ,where t_{final}=time from the moment of the throw until the stone reaches the climax of the trajectory...

    Tell me if you get zero in this case...

  7. Feb 27, 2005 #6
    Indeed, i made a mistake there. However i think the best way tosolve this is like this (i don't see another way out)

    g = 9.81

    so the second formula gives us

    1.14 = v_0*0.2 - 9.81*0.2²/2 Solve this for V_0

    Then to determin the maximal heigth

    v = 0 = v_0 (which you know from above calculation) -9.81t Solve for t and plug this t-value into the second formula for y...Now you know the maximal height...then just subtract 1.14

  8. Feb 27, 2005 #7
    Yes you will get ZERO, because in your attempt to correct me, you made a mistake yourself. You shouldn't have written v as a VECTOR. You will get ZERO because the initial and final vector for v have OPPOSITE direction and are equal in length...

    To the OP : disregard dexter's post because it is both wrong and useless in this case

  9. Feb 27, 2005 #8


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    Lol...The final velocity is zero (did u read my post...?Why do you make gratuitious affirmations...?),because i chosed the t_{final} to be the moment of time at which the stone is at its climax,where the velovity is zero...



    P.S.To the OP:disregard all posts...Mine & Marlon's included...:tongue2:
  10. Feb 27, 2005 #9
    The final velocity is NOT zero. Have you forgotten about energy conservation. The final velocity is the velocity when the object is back on the ground. This is where you went wrong.

    However this is not relevant to this problem. I posted a possible way out using the formula's for y and v


    now i am off, the academy awards show has begun...
    Last edited: Feb 27, 2005
  11. Feb 27, 2005 #10

    well, i thank everyone for their help in this physics problem, however "disregard all posts...Mine & Marlon's included..." stated by dextercioby is confusing me as i still do not know where to start in solving this problem. pleez, someone confirm a post that would definitely assist me and come to a proper answer.

  12. Feb 27, 2005 #11
    Jen, just look at posts 2 and 6...i practically gave the answer there

    now i am really gone
  13. Feb 27, 2005 #12
    Thanks Marlon. I am now aware that the answer to the problem is 1.8m. if anyone received a different answer, please feel free to explain.
  14. Feb 27, 2005 #13


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    Yes,the answer is (weirdly) correct...:approve:

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