Solving for Total Distance on 9th Bounce: Help Needed!

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In summary: Thank you HallsofIvyIn summary, the basketball is dropped from a height of 15 m and it bounces to 70% of the previous height. The total distance the ball travels when it hits the ground for the ninth time is 48.0 cm.
  • #1
Mspike6
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I have always had problems with Swquences and sum :S

A basketball is dropped from a height of 15 m and bounces to 70% of the previous height. The total distance the ball travels when it hits the ground for the ninth time is (give answer to the nearest tenth)


i know we probably going to use sn=a(1-2n)/1-r somewhere.
and we going to use Tn-1 (0.70) ... but i really can't get it right.
 
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  • #2
Nvm, i think i got it..

sn=a(1-r^n)/1-r
sn=15(1-(0.70^9) / (1-0.70)

sn=48.0But there is anther one that am having troubles with

An oil well produces 25 000 barrels of oil during its first month of production. The oil company predicts its production will drop 7% each month thereafter. How many barrels of oil will this company produce in its first year? Round your answer to the nearest thousand.

Edit:
i could get that Tn= Tn-1 * 0.93
 
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  • #3
Mspike6 said:
Nvm, i think i got it..

sn=a(1-r^n)/1-r
sn=15(1-(0.70^9) / (1-0.70)

sn=48.0


But there is anther one that am having troubles with

An oil well produces 25 000 barrels of oil during its first month of production. The oil company predicts its production will drop 7% each month thereafter. How many barrels of oil will this company produce in its first year? Round your answer to the nearest thousand.

Edit:
i could get that Tn= Tn-1 * 0.93
Yes, your sum is 25000+ 25000(.93)+ 2500(.93)^2+ ... which is
[tex]\sum_{n=0}^{11}25000 (.93)^n[/tex]
(n goes up to 11 because there are 12 months in a year and we started the numbering at "0".)

That sum is, using the same formula you did before,
[tex]25000\frac{1- (.93)^{12}}{1- .93}[/tex]
 
  • #4
Thank you HallsofIvy

I got this new QUestion about Lim.

Lim(x-->0) Sinx/Tan X

I think it will go something like.

Lim(x-->0) Sinx / (Sinx/Cosx)

Lim(x-->0) Sinx * Cosx/Sinx

Then am stuck from here :P Thanks guysm, your help is REALLY appreciated.and btw. is there a good Tutorial on Treg. On general or Specificly on Treg Limits ?
 
  • #5
Mspike6 said:
Thank you HallsofIvy

I got this new QUestion about Lim.

Lim(x-->0) Sinx/Tan X

I think it will go something like.

Lim(x-->0) Sinx / (Sinx/Cosx)

Lim(x-->0) Sinx * Cosx/Sinx

Then am stuck from here :P

You can now divide sin x by sin x which will leave you with 1*cos x. what happens when x-->0 now?
 
  • #6
Thanks Furrygoat !..

It will be 1.

Thanks again :D
 
  • #7
Mspike6 said:
Thank you HallsofIvy

I got this new QUestion about Lim.

Lim(x-->0) Sinx/Tan X

I think it will go something like.

Lim(x-->0) Sinx / (Sinx/Cosx)

Lim(x-->0) Sinx * Cosx/Sinx
So it is like "(ab)/a". As long as a is not 0, you can cancel: (ab)/a= b.
As long as sin(x) is not 0 (that is, as long as x is not 0) sin(x)cos(x)/sin(x)= cos(x).

Important "law of limits" that is often overlooked: if f(x)= g(x) for all x except x= a then [itex]\lim_{x\to a} f(x)= \lim_{x\to a} g(x)[/itex].

Then am stuck from here :P


Thanks guysm, your help is REALLY appreciated.


and btw. is there a good Tutorial on Treg. On general or Specificly on Treg Limits ?
 

Related to Solving for Total Distance on 9th Bounce: Help Needed!

1. What is the formula for calculating total distance on the 9th bounce?

The formula for calculating total distance on the 9th bounce is: D = (2^(n-1)) * D0, where D is the total distance traveled, n is the number of bounces, and D0 is the initial distance of the first bounce.

2. How do I know what value to use for n in the formula?

The value of n in the formula represents the number of bounces. So if you want to solve for the total distance on the 9th bounce, n would be equal to 9.

3. Can I use this formula for any type of bouncing object?

Yes, this formula can be used for any type of bouncing object as long as the initial distance and the height of the bounces remain constant.

4. What units should I use for the initial distance and total distance?

The units used for the initial distance and total distance will depend on the units used for the height of each bounce. For example, if the height of each bounce is measured in meters, then the units for the initial distance and total distance will also be in meters.

5. Is there a limit to how many bounces this formula can be used for?

No, there is no limit to how many bounces this formula can be used for. It can be used for any number of bounces, as long as the initial distance and height of the bounces remain constant.

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