A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of $4\mathrm{.}91$ m$/$s$.$ The rock misses the edge of the cliff as it falls back to earth. If the rocks hits the bottom of the valley, which is $44\mathrm{.}43$ meters below the cliff, what is the speed of the rock when it lands, in unit of m$/$s$?$ Neglecting the effects of air resistance. g$=9\mathrm{.}8$ m$/$s$2.$