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Bouncing ball math question

  1. Jun 9, 2005 #1
    This is a pretty tricky question...been trying to piece it together but I think I need some help.

    A ball is dropped from a height of 3 m. After each bounce, it rises to 82% of its previous height. After how many bounces does the ball reach a height less than 15cm?

    I get the whole geometric series thing, finding 'n' and all that, but what's with the "less than 15cm"? What do I do in this case? :confused:
     
  2. jcsd
  3. Jun 9, 2005 #2
    If initial height = h_i = 3m
    Height of nth bounce = h_n = 3*.82^n

    You are looking for n such that h_n < 0.15m.
     
  4. Jun 9, 2005 #3
    Exactly so would I just choose any value <0.15m and use it as h_n in order to find 'n'?
     
  5. Jun 9, 2005 #4
    That wouldn't work, because the series h_n only holds specific values. I'd just crank out values for h_n for each n, my guess is its about 15.
     
  6. Jun 9, 2005 #5
    So you mean kind of like a trial and error approach?
     
  7. Jun 9, 2005 #6
    Height initially = [itex]h[/itex] = 3m
    Height after first bounce = [itex](0.82)(h)[/itex]
    height after second bounce=[itex](0.82)^2 (h)[/itex]
    Height after nth bounce=[itex](0.82)^n(h)[/itex]

    Now here you might like to use a bit of 'hitntrial' ...Now you want 'n' such that [itex](0.82)^n(h)[/itex] is just more than .15m
     
  8. Jun 9, 2005 #7
    Why more if the question specifies that it should be less?
     
  9. Jun 9, 2005 #8
    That's what I would do, but I'm a lazy ass.
     
  10. Jun 9, 2005 #9
    It should be less.
     
  11. Jun 9, 2005 #10
    [itex](0.82)^n>0.05[/itex]

    [itex]
    n=15
    [/itex]
     
  12. Jun 9, 2005 #11
    Brain, that is just confusing, let her solve the problem her own way. Ms. confused,

    [tex] h_n = 3(0.84)^n [/tex] and [tex] h_n \leq 0.15 [/tex] so then

    [tex] 3(0.82)^n \leq 0.15 [/tex]

    You need to solve that for an integer n.
     
    Last edited: Jun 9, 2005
  13. Jun 9, 2005 #12
    actually i misread the question .

    n=15 gives the exact answer,but for less than 15 cm , n=14
     
  14. Jun 9, 2005 #13
    Okay, I was getting 15 too but according to the answer key, n=16. This is probably a type-o, eh?
     
  15. Jun 9, 2005 #14
    No, 16 is the correct solution.

    Brain, the sequence is decreasing, for each higher n, h_n has a lower value. At n = 15 you get

    [tex] 3*0.82^{15} = 0.1528m [/tex] which is still higher than 0.15m. Therefore you need another (16th) bounce to drop below.
     
  16. Jun 9, 2005 #15
    Thanks for correction.n=16
     
  17. Jun 9, 2005 #16
    Alright now it makes perfect sense. Thank you so much!
     
  18. Jun 9, 2005 #17
    n= 16 because,
    log0,05/log0,82=15,... at this value of n, h is 15 cm. on 16th jump, it would get under that height.
     
  19. Jun 9, 2005 #18

    Curious3141

    User Avatar
    Homework Helper

    The much easier way to solve the inequality is to use logs.

    Let [itex]N[/itex] be the required number of bounces, [itex]h_0[/itex] be initial height in centimeters and [itex]h_n[/itex] be the height after [itex]n[/itex] bounces.

    We have : [tex]h_n = (0.82)^nh_0[/tex]

    We want : [tex]h_N < 15[/tex]

    So [tex](0.82)^Nh_0 < 15[/tex]

    Take common logs (base 10) of both sides,

    [tex]\log{(0.82)^N} + \log h_0 < \log 15[/tex]

    The above inequality holds because log is a monotone increasing function on the positive reals.

    [tex]N\log(0.82) < \log 15 - \log h_0[/tex]

    Divide throughout by [tex]\log(0.82)[/tex], but remember to invert the sign of the inequality because this is a negative quantity (since 0.82 is less than 10, the base of the logs).

    [tex]N > \frac{\log 15 - \log h_0}{\log(0.82)}[/tex]

    Evaluate that, substituting [tex]h_0[/tex] = 300,

    giving [tex]N > 15.096[/tex] or so.

    So the smallest integer value for N that satisifies is [tex]N = 16[/tex].
     
  20. Jun 10, 2005 #19
    I think the poster's problem was the setup, not the solving.
     
  21. Jun 10, 2005 #20

    Ouabache

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    Science Advisor
    Homework Helper

    As you found out on iteration [tex] 3*0.82^{15} = 0.1528 [/tex]. This is better to use, compared to [tex] 3*0.82^{15} = 0.15 [/tex]
    This is where those rounding errors, I mentioned on an earlier question, come into play.

    Doc Brain, I liked your approach to solving this one.
    Another way to plug-and-chug an interative formula is using spreadsheet. Excel's math utilities handle calculations well.
     
    Last edited: Jun 11, 2005
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