time to fall from the table..
-1.0= - 1/2 g t^2
or t= you do it.
Now it went .25 m horizontally in t seconds
vtableedge*time=.25m
solve for vtableat edge.
Now one can look at the table top.
Vf^2=Vi^2+2ad where a is frictionforce/mass 0r
a=-mg*mu/m or -mu*g
vf^2=vi^2 -2*mu/g * 2.0
solve for vi
A popular pastime is to see who can push an object closest to the edge of a table without its going over. You push a 130 g object and release it 2.00 m from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge, falls 1.00 m to the floor, and lands 0.25 m from the edge of the table. If the coefficient of kinetic friction between the object and the table is 0.55, what was the object's speed (in m/s) as you released it, assuming that the local acceleration due to gravity is -9.80 m/s2?
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