Math help please?

[G3CK0]

Hey there. I got a little homework problem and I hope that someone can help me out 

The question is: If a rubber ball can bounce back to 90% of its original height, what is the total vertical distance that it will travel if it is dropped from an elevation of 40 feet?

Thanks a lot.

[lucid]

76ft

[Z3R0]

76ft

@OP, I concur with lucid's deduction. Granted that we are calculating the total distance traveled from one bounce between two inert points of starting and stopping.

[lucid]

That's what I'm assuming. Otherwise it's 36ft.

[G3CK0]

You guys are a great help <3.

But the thing is that I have to show my work, so is there a specific formula where you got the result from or do I just write the result straight ahead?

[lucid]

What the question is basically asking is; What is ninety percent of forty, plus forty.

I didn't really use a formula other than that. I just did it in my head.

[G3CK0]

Why's it plus 40 after the percentage?

[lucid]

If I understood the question right you are asking how far does it travel from where you dropped it(40ft up) to it hitting the ground and bouncing 90 percent back up. So that's forty feet down, and thirty six feet up.

[G3CK0]

Ah, thank you very much for your time.

[s3my0n]

Some working out:

s = 40 f
S(total) = ?

S(total) = s + 0.9s = 1.9s = 1.9 * 40 = 76 f

[p_2001]

No.. The ball does not stop after one bounce..
after first bounce it will rise to 36.ft.. Then ninety percent of 36 ft... And so on... The royal well be like..

40+36+36+36*.9 + 36*.9 + 36*.9*.9 + 36*.9*.9 and so on..
It will be a GP.. And you need to apply infinite gp..
Find the formula for summation yourself.

[s3my0n]

No.. The ball does not stop after one bounce..
after first bounce it will rise to 36.ft.. Then ninety percent of 36 ft... And so on... The royal well be like..

40+36+36+36*.9 + 36*.9 + 36*.9*.9 + 36*.9*.9 and so on..
It will be a GP.. And you need to apply infinite gp..
Find the formula for summation yourself.

True that, I got misdirected by the previous answers x]

Then it would be:


Where:
x = original height
k = percentage change each bounce (%/100)


UPDATE:


The above formula is the same as:



And Newton found that that will equal to:



So the answer to the problem is 40/(1-0.9) = 400 feet

[p_2001]

True that, I got misdirected by the previous answers x]

Then it would be:


Where:
x = original height
k = percentage change each bounce (%/100)


UPDATE:


The above formula is the same as:



And Newton found that that will equal to:



So the answer to the problem is 40/(1-0.9) = 400 feet

No, a bit more tricky..
See.. The ball travels TWICE .. Read what i wrote..once it goes up and then the same distance down...
So it will be..
400+400-40... = 760..

Also i would say it is hw help.. No need to give out formulas here .. Let the op do some work.

[lucid]

I considered this but I assumed the question was asking how far in one bounce.

[p_2001]

I considered this but I assumed the question was asking how far in one bounce.

The Question said "total vertical distance"..
It said ball is dropped and hence only dropped not picked or caught after one bounce.