If Upton is falling from rest (i.e. from the max height of the ride), then his (?) velocity is at time [tex]t[/tex] is given by
[tex]v=gt[/tex]
where [tex]g=9.80\,\frac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. Note that I'm taking the downward direction to be positive, and hence the acceleration also has positive direction. After 2.60 seconds, Upton's velocity will be
[tex]v=\left(9.80\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.60\,\mathrm s)=25.48\,\dfrac{\mathrm m}{\mathrm s}\approx25.5\,\dfrac{\mathrm m}{\mathrm s}[/tex]
(rounding to 3 significant digits)
