Chapter 9.2, Problem 73E

To determine

Find the distance in feet when the object fall in the first 7 second after it is dropped.

Answer to Problem 73E

In seven seconds the object falls a distance of 784 ft..

Explanation of Solution

Given:

The object fall in 1st second is 16ft

Object fall in 2nd second is 48ft

Object fall in 3rd second is 80ft

In 4th second it fall 112 ft

  $a_1=16,d=32$

The object fall in 1st second is 16ft

Object fall in 2nd second is 48ft

Object fall in 3rd second is 80ft

In 4th second it fall 112 ft

Thus the terms $16,48,80,112$ ,....form an arithmetic sequence with common difference 32

Here $a_1=16,d=32$

The sum of n term of the arithmetic sequence is

  $S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$ .....(1)

Let $a_1=16,d=32,n=7$ in (1) then

  $\begin{array}{l}\begin{array}{l}{S}_n=\frac{7}{2}\left(2\left(16\right)+\left(7–1\right)32\right)\hfill \\ =\frac{7}{2}\left(7\right)\left(32\right)\hfill \end{array}\\ =784\end{array}$

Hence in seven seconds the object falls a distance of 784 ft.