Question 76190
the position (s) of an object moving under constant acceleration (a) with initial velocity (v) and initial position (p) in time (t) is {{{s=(at^2)/2+vt+p}}}


using 32ft/sec/sec as the gravitational acceleration {{{s=-16t^2+20t+.5}}}...the first term is negative because gravity acts downward


setting s=0 (cork on table) and solving for t gives...{{{t=(-20 +- sqrt(20^2-4*(-16)*.5))/(2*(-16))}}}...or {{{t=(-20 +- sqrt(432))/(-32)}}}...or {{{t=.625 +- .375 (sqrt(3))}}}


so t approximately equals 1.27 sec.