81 m/s²), and $s$ is the height (1.95 m). Plugging in the values, we get:
$v^2 = 0^2 + 2(9.81)(1.95)$
$v^2 = 38.229$
$v = \sqrt{38.229} = -6.18$ m/s (negative sign indicates downward direction)
So, the velocity just before it strikes the floor is $\boxed{-6.18
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