College Algebra, Chapter 9, 9.2, Section 9.2, Problem 66

When an object is allowed to fall freely near the surface of the earth, the gravitational pull is such that the object falls 16 ft in the first second, 48 ft in the next second, 80 ft in the next second, and so on.

a.) Find the total distance a ball falls in 6 s.

b.) Find a formula for the total distance a ball falls in $n$ seconds.

a.) If we let $D_T$ be the total distance of the ball after 6 s, then


$\begin{equation} \begin{aligned} D_T =& 16 + 48 + 80 + 112 + 144 + 176 \\ \\ D_T =& 576 \text{ ft} \end{aligned} \end{equation}$


b.) If $a = 16$ and $a_2 = 48$, then their common difference $d = 32$. Therefore, the total distance the ball fell in $n$ seconds is represented as..


$\begin{equation} \begin{aligned} D_n =& \frac{n}{2} [2(16) + (n-1)(32)] \\ \\ D_n =& \frac{n}{2} [32 + 32(n-1)] \\ \\ D_n =& \left( \frac{n}{2} \right) (32) [1 + (n-1)] \\ \\ D_n =& 16n [n] \\ \\ D_n =& 16n^2 \end{aligned} \end{equation}$