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Homework Help: Time and height in a freefall problem

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A marble fell from the height h and covered the last 196,2m in 4 seconds. How long was the marble falling? What height was it falling from?

    2. Relevant equations

    [itex]v(t) = v_0 + at[/itex]

    3. The attempt at a solution

    In the first equation above, we can plug what we know, obtaining: [itex]196,2=v_{0}\cdot4+\frac{9,81\cdot4^{2}}{2}=4v_{0}+78,48[/itex] so [itex]v_0 \approx 29,4m/s[/itex]. We can plug it into the second obtaining [itex]v(4)=29,4+9,81\cdot4=68,64m/s[/itex]. What does it give me, though? How should I progress?
  2. jcsd
  3. Oct 15, 2012 #2
    Could you please help me how to go on?
  4. Oct 15, 2012 #3
    You have the initial velocity at the beginning of the last 4 seconds. But this initial velocity is NOT a function of 4 seconds, it is a function of unknown time BEFORE those 4 seconds. Use the equation to obtain this unknown time.
  5. Oct 15, 2012 #4
    s = Δs + gt²/2 = 196.2
    you can find the speed at t= t-4

    v = g(t-4)
    this is is your initial velocity at the height H-196.2

    so Δs = v(t-4) = g(t-4)(t-4) = g(t-4)²

    g(t-4)² + g(t-4)²/2 = 196.2

    or the more convinient way, which I like to use:

    2g(t-4)² + g(t-4)² = 2 * 196.2 or 392.4
    3g(t-4)² = 392.4

    So you have a standard quadratic equation and the rest ought to be pretty simple I think.
    So if you solve for t, you get the entire time the marble was freefalling.
  6. Oct 16, 2012 #5
    Thank you. Using this, though, it turns out t=4s. Does that mean that the entire freefall was the 4s mentioned in the task specification?
  7. Oct 16, 2012 #6
    No that is not what I got as my answer. We are solving the same equation, perhaps you d tell me what your way of going about it was.

    Anyway, another thing, if 4s were the entire time of the free fall then it would be contradicting the freefall distance equation which says that your maximum distance in 4s without initial acceleration would be - 78.4m
  8. Oct 16, 2012 #7
    Oh, you are right, what a silly mistake in my equation :) So we have t=7.68s, right? Then, we have 3,68s which was not calculated in the task during which it covers gt²/2=66m so summing up, it was falling from a height equal to 66+196.2=262.2m, is that OK?
  9. Oct 16, 2012 #8
    Slight difference from mine due to the rounding, but yes, that is the correct time :)
  10. Oct 16, 2012 #9
    Thank you :) So does the second part of my post also make sense?
  11. Oct 16, 2012 #10
    I apologise for mis leading you - I think things in my own head weren't moving in the correct direction when I thought of the solution I gave you before.

    The distance the marble travels during t-4 is not known and the total distance travelled during the t seconds is gt²/2
    and now you get
    g(t-4)²/2 + 196.2 = gt²/2
    I solved it through and this time the numbers do add up.
    For some reason I had assumed before that t-4 seconds is the time for the last 196m. My mind must have been in freefall or something.
    this makes much more sense than the last one. I apologise again for mis-leading you, but I guess if I fix the problem, in the end we're all pals, ey? :)
    Last edited: Oct 16, 2012
  12. Oct 16, 2012 #11
    Haha, we sure are, thank you so much ;)
  13. Oct 16, 2012 #12
    I corrected the above equation as follows

    s = ut + gt2/2

    u = g (t0)
    t = 4

    so 196.2 = 9.81 * t0 * 4 + 9.81 * 4 * 4 / 2
    t0 = 3

    The marple fell 3 seconds to cross 44.15 m, then it travelled 4 seconds and covered 196.2 m

    So the marple fell 240.35 m in 7 seconds.
  14. Oct 17, 2012 #13
    Yes, thank you, I can't even imagine where I got my idea for the equation I got to at first. My corrected version follows the same logic as yours, icetiger. Making mistakes is human :P
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