you have to use the kinematics equation s= u*t - 0.5gt^2 in vertical and horizontal directions. u is the initial velocity and s is the distance travelled. g is the gravi . acc = 9.8 ms-2 , when it reaches floor level it has traveled -1m from the place of launch. use above equation for vertical direction with s= -1m , and u=7.66* sin30 , you have to resolve for the vertical component of launch speed, now when you plug in this data in above equation you will get a quadratic equation in t you can find two roots to this equation , one root will be (-) so you can take the positive root, this is the time it takes to get to the floor level, now think of the horizontal displacement there is no acceleration in that directin so the resolved horizontal speed ( 7.66*cos 30) will remain unchanged so the horizontal dispacement = horizontal speed * time it takes to getto the floor !
good luck