A ball is projected horizontally from the edge of a table that is 0.823 m high, and it strikes the floor at a point 1.44 m from the base of the table. The acceleration of gravity is 9.8 m/s2 . What is the initial speed of the ball? An- swer in units of m/s. the table is .823 m high. x = xo + vot+ .5at^2 x = .823 , xo = 0, vot = 0 (vertical speed is 0), a = 9.8 m/s2 and you solve for t .823 m = .5(9.8 m/s2)t^2 t^2 = .1679 s t = .41 sec to find the intial speed you know the distance and the time x = xo +vot+.5at^2 1.44 m = 0 m + vo(.41 s) + .5(9.8 m/s2)(.41 s)^2 you get vo = 1.5 m/s I tried this but I did something wrong because i enter my hw online and it told me the answer i got was incorrect.