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A ball is dropped from a height of 6 feet. With each bounce, the ball rebounds to a height that is three-quarters as high as the previous bounce. Represent the height to which the ball rises after it hits the floor for the nth time using a recursive formula and an explicit formula. Recursive Formulas: anar.an-1 or an+1 =rean Explicit Formula: an = Q .m-1 Represent (using sigma notation) the total vertical distance the ball has traveled when it hits the floor for the fifth time. Calculate this distance. ? Σ ? k=? Sn 4,(1-r") 1-r

A ball is dropped from a height of 6 feet. With each bounce, the ball rebounds to-example-1
User Jtrick
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1 Answer

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We know that each time the ball bounces it travels a distance of 3/4 the original height, that means that for each bounce the ball will rise:


a_n=(3)/(4)a_(n-1)

where a_n is the height of the nth bounce and a_(n-1) is the bounce n-1 (the previous one).

Now, from this formula we know that:


a_1=(3)/(4)a_0=(3)/(4)\cdot6=(9)/(2)

Therefore the explicit formula for the height of the nth bounce is:


\begin{gathered} a_n=(9)/(2)((3)/(4))^(n-1) \\ =(9)/(2)\cdot(4)/(3)((3)/(4))^n \\ =6((3)/(4))^n \end{gathered}

Then:


a_n=6((3)/(4))^n

Now, to find the total distance it travels by the fifth bounce we follow:

The first time the ball hits the ground it has traveled a distance of 6 ft.

Now, for the second bounce we have:


6((3)/(4))+6((3)/(4))=12((3)/(4))

(one distance up and one distance down).

For the third bounce we have:


6((3)/(4))((3)/(4))+6((3)/(4))((3)/(4))=12((3)/(4))^2

If we continue this process we notice that the ball will travel a total distance of:


6+12((3)/(4))+12((3)/(4))^2+12((3)/(4))^4+\ldots\text{..}

but this can be written (for an infinity number of bounces) as:


\begin{gathered} 6+12\sum ^(\infty)_(n\mathop=0)((3)/(4))^(n+1) \\ =6+12((3)/(4))\sum ^(\infty)_(n\mathop=0)((3)/(4))^n \\ =6+9\sum ^(\infty)_(n\mathop=0)((3)/(4))^n \end{gathered}

As we said this is for an infinity of bounces if we like only five bounces then we have:


6+9\sum ^5_(n\mathop=0)((3)/(4))^n

Now to find the distance in this case we do the sum and we get appoximately 35.5928 feet.

User Peter Stephens
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