The velocity \(v\), in meters per second, of a falling object on Earth after \(t\) seconds, ignoring the effect of air resistance, is modeled by the equation \(v=9.8t\). There is a different linear relationship between time and velocity on Mercury, as shown in the table below.

$$ \begin{array}{|c|c|} \hline \text{Time} & \text{ Velocity on} \\ \text{ (seconds)} & \text{Mercury (meters} \\ & \text{ per second)} \\ \hline 0 & 0 \\ \hline 2 & 7.4 \\ \hline 4 & 14.8 \\ \hline \end{array} $$

If an object dropped towards the surface of Earth has a velocity of 117.6 meters per second after \(t\) seconds, what would be the velocity of the same object dropped toward the surface of Mercury after \(t\) seconds, ignoring the effect of air resistance?

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Let's start by finding the time \(t\) given the condition that \(v=117.6\) on Earth.

$$ \text{velocity on earth}=9.8t $$ $$ 117.6=9.8t$$ $$ t=12 $$

From the table, an object appears to achieve a velocity of 7.4 meters per second after 2 seconds. This is equivalent to 3.7 meters per second after 1 second.

$$ \text{velocity on mercury}=3.7t $$ $$ v=3.7(12) $$ $$ v=\boxed{44.4 \text{ meters per second}}$$