Bouncing ball math question

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:
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.
Exactly so would I just choose any value <0.15m and use it as h_n in order to find 'n'?
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.
So you mean kind of like a trial and error approach?
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
Why more if the question specifies that it should be less?
That's what I would do, but I'm a lazy ass.
It should be less.

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.
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actually i misread the question .

n=15 gives the exact answer,but for less than 15 cm , n=14
Okay, I was getting 15 too but according to the answer key, n=16. This is probably a type-o, eh?
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.
Thanks for correction.n=16
Alright now it makes perfect sense. Thank you so much!
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.


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].
I think the poster's problem was the setup, not the solving.


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.
Dr. Brain said:
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]
Another way to plug-and-chug an interative formula is using spreadsheet. Excel's math utilities handle calculations well.
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