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 $1-.8=.2$)
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 $80\%*y$.