How many bounces until the ball reaches a height less than 15cm?

[ms. confused]

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?

 

[whozum]

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.

 

[ms. confused]

Exactly so would I just choose any value <0.15m and use it as h_n in order to find 'n'?

 

[whozum]

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.

 

[ms. confused]

So you mean kind of like a trial and error approach?

 

[Dr.Brain]

Height initially = $h$ = 3m
Height after first bounce = $(0.82)(h)$
height after second bounce=$(0.82)^2 (h)$
Height after nth bounce=$(0.82)^n(h)$

Now here you might like to use a bit of 'hitntrial' ...Now you want 'n' such that $(0.82)^n(h)$ is just more than .15m

 

[ms. confused]

Why more if the question specifies that it should be less?

 

[whozum]

That's what I would do, but I'm a lazy ass.

 

[whozum]

It should be less.

 

[Dr.Brain]

$(0.82)^n>0.05$

$n=15$

 

[whozum]

Brain, that is just confusing, let her solve the problem her own way. Ms. confused,

$$h_n = 3(0.84)^n$$ and $$h_n \leq 0.15$$ so then

$$3(0.82)^n \leq 0.15$$

You need to solve that for an integer n.

 

[Dr.Brain]

actually i misread the question .

n=15 gives the exact answer,but for less than 15 cm , n=14

 

[ms. confused]

Okay, I was getting 15 too but according to the answer key, n=16. This is probably a type-o, eh?

 

[whozum]

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

$$3*0.82^{15} = 0.1528m$$ which is still higher than 0.15m. Therefore you need another (16th) bounce to drop below.

 

[Dr.Brain]

Thanks for correction.n=16

 

[ms. confused]

Alright now it makes perfect sense. Thank you so much!

 

[wisredz]

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.

 

[Curious3141]

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

Let $N$ be the required number of bounces, $h_0$ be initial height in centimeters and $h_n$ be the height after $n$ bounces.

We have : $$h_n = (0.82)^nh_0$$

We want : $$h_N < 15$$

So $$(0.82)^Nh_0 < 15$$

Take common logs (base 10) of both sides,

$$\log{(0.82)^N} + \log h_0 < \log 15$$

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

$$N\log(0.82) < \log 15 - \log h_0$$

Divide throughout by $$\log(0.82)$$, 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).

$$N > \frac{\log 15 - \log h_0}{\log(0.82)}$$

Evaluate that, substituting $$h_0$$ = 300,

giving $$N > 15.096$$ or so.

So the smallest integer value for N that satisifies is $$N = 16$$.

 

[whozum]

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

 

[Ouabache]

As you found out on iteration $$3*0.82^{15} = 0.1528$$. This is better to use, compared to $$3*0.82^{15} = 0.15$$
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.

Height initially = $h$ = 3m
Height after first bounce = $(0.82)(h)$
height after second bounce= $(0.82)^2 (h)$
Height after nth bounce=$(0.82)^n(h)$

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

 

[whozum]

Brain's was wrong