For a distance, #d#, an initial velocity, #v_0#, an acceleration of #a#, and a time of #t#
#color(white)("XXX")d=v_0t+at^2#
In this case
#color(white)("XXX")d=-2 " m"# (2 meters below the height from with it was thrown)If it was dropped and not thrown up there would be no negative sign
#color(white)("XXX")v_0= 2 " m"/"sec"#
#color(white)("XXX")a=-9.8 " m"/(sec"^2)# (standard gravity assumed)
So we have
#color(white)("XXX")-2.0" m" = 2.0t " m" + (-9.8t^2)" m"#
#color(white)("XXX")98t^2-20t-20=0#
Using the quadratic formula:
#color(white)("XXX")t=(20+-sqrt(20^2-4(98)(-20)))/(2(98))#
#color(white)("XXX")t = 0.56518s#