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.

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