Chapter 8.2, Problem 86E

To determine

To find: the height from which the object will fall in 8 seconds.

Answer to Problem 86E

Object falls 1024 feet in 8 seconds.

Explanation of Solution

Given information:

During the first second of fall, the object falls 16 feet; during the second, it falls 48 feet; during the third second, it falls 80 feet; and during the fourth second, it falls 112 feet.

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}$

Common difference can be defined by d .

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}=d$

nth term of the arithmetic sequence has the form

  $a_n=a_1+\left(n-1\right)d$

Where $a_1$ is the first term of the sequence, and d is the common difference.

Sum of an arithmetic finite sequence has the form

  $S_n=\frac{n}{2}\left(a_1+a_n\right)$

Here, n is number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last term of the sequence.

Calculation:

First second of object falls $a_1=16$

Second of object falls $a_2=48$

The common difference between these seats is

  $\begin{array}{c}d=a_2-a_1\\ ==48-16\\ =32\end{array}$

The objective falls in 8 second by using formula, $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

  $\begin{array}{c}{S}_8=\frac{8}{2}\left(2\times 16+\left(8-1\right)32\right)\\ =4\left(32+224\right)\\ =4\times 256\\ =1024\end{array}$

Hence object falls in 8 seconds is 1024.