Question

The distance that a free falling object falls is directly proportional to the square of the time it falls (before it hits the ground). If an object fell 89ft in 2 seconds, how far will it have fallen by the end of 8 seconds? (Leave the variation constant in fraction form or round to at least 2 decimal places. Round your final answer to the nearest foot.)

Answer 1

Given:

The object fell 89 ft.

Time is taken 2 sec.

Find:

How far in 8 sec.

Sol:

Distance, d varies directly with the square of the time, t.

$\begin{gathered} d\propto t^2 \\ \\ d=kt^2 \end{gathered}$

If d = 89 and t=2

$\begin{gathered} d=kt^2 \\ \\ 89=k(2)^2 \\ \\ k=\frac{89}{4} \\ \\ k=22.25 \end{gathered}$

So the equation is:

$d=22.25t^2$

Distance after 8 sec.

$\begin{gathered} d=22.25t^2 \\ \\ d=22.25(8)^2 \\ \\ d=22.25\times64 \\ \\ d=1424 \end{gathered}$

So fell after 8 seconds is 1424 ft