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A ball falls from a height of $2$ meters onto a firm surface and jumps after each impact each back to $80\%$ of the height from which it fell. ¨ Set up the function, which indicates the height of the ball after the $n$th impact reached. How high does the ball jump after the $5$th impact?

Below question is considered as a linear growth and not exponential growth . I dont understand why its linear growth or decay. $$y=2*0.8^5$$

For the percentage neither it add to $1$ nor minus from $1$.

I wish to ask why it's not $y=2*.2^5$ (I took $80\%$ as decay and did $1-.8=.2$)

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Here's what you've done.

With each impact, the height of the bounce decreases to $80\%$ of what it was. But you've made it decrease by $80\%$ (to $20\%$) by subtracting the $80\%$ from $1$.

I think that's where your confusion is.

You need to multiply the height by $80\%$ with each bounce, and the decay comes from the fact that $80\%$ is smaller than $1$.

Now, looking at it step by step:

After no bounces (the starting position), you've multiplied by $0.8$ no times, so the height in metres is $$y=2=2*0.8^0$$

($0.8^0$ simply means "Don't multiply by $0.8$ at all": anything to the power of $0$ is $1$. But putting it in helps to show the pattern.)

After $1$ bounce, you've multiplied by $0.8$ once, so $$y=2*0.8=2*0.8^1$$

After $2$ bounces, you've multiplied by $0.8$ twice, so now $$y=2*0.8*0.8=2*0.8^2$$

After $3$ bounces, you've multiplied by $0.8$ $3$ times, making $$y=2*0.8*0.8*0.8=2*0.8^3$$

After $n$ bounces, you've multiplied by $0.8$ $n$ times, so $$y=2*0.8^n$$ which is the equation the question wants you to use.

Then just put $n=5$ to get the height after $5$ bounces.

Edit: When a problem like this is described in words, "of" is very often the same as "multiplied by": e.g. "two thirds of $x$" means $\frac23*x$ and "$80\%$ of $y$" means $80\%*y$.

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  • $\begingroup$ My question is why its linear growth and not an exponential growth $\endgroup$ – tomtom Dec 16 '18 at 8:56
  • $\begingroup$ It's not linear—that would mean adding or subtracting a fixed amount with each bounce. It's exponential because we're multiplying by a fixed amount ($0.8$) each time, and it's decay because repeatedly multiplying something by $0.8$ makes it get smaller and smaller. If we were multiplying by something bigger than $1$ (e. g. $1.1$ in $y=a*1.1^n$) then it would be growth. $\endgroup$ – timtfj Dec 16 '18 at 11:40
  • $\begingroup$ i am quite lost . eg the value of a new car in 2015 is 4000. it depreciates 7% every year. how much will the car be worth in 2024. here i notice that solution is first find (1-0.07)= 0.93 and then 4000(0.93)^9 is the answer. may i ask why some time for decay we minus decaly percentage from 1 and in above we just dont . may i ask why $\endgroup$ – tomtom Dec 16 '18 at 20:03
  • $\begingroup$ If it depreciates by $7\%$, it goes down to $93\%$ of what it was. So, multiply by $0.93$, $9$ times (for the $9$ years): $4000*0.93^9=2081.64$ $\endgroup$ – timtfj Dec 16 '18 at 20:13
  • $\begingroup$ It's the difference between decreasing by a certain amount (which needs subtracting from $100\%$) and decreasing to a certain amount. The key thing is to work out what multiple the new value is of the old one. $\endgroup$ – timtfj Dec 16 '18 at 20:17

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