Question 117873
First we need to find out {{{how}}}{{{ high}}} the ball will go {{{before}}} it starts falling back to Earth.


{{{vf = v[o] + at}}}


{{{0 = 20 - 9.81t}}}


{{{9.81t = 20}}}


{{{t = 20/(9.81)}}}


{{{t = (2.04) s}}}


{{{s = v[o]t + (1/2)at^2}}}



{{{s = 20(2.04) + (1/2)(-9.81)(2.04)^2}}}


{{{s = 20(2.04) + 4.9*4.16}}}


{{{s = 40.8 - 20.4}}}


{{{s = (20.4) m}}}


Now {{{to}}}{{{ get}}}{{{ to}}}{{{ 80m}}} it must travel back the {{{20.4m}}} to the top of the roof and{{{ then}}}{{{ fall}}} another {{{20 m}}} for a total of {{{(40.4) m}}}


{{{s = (1/2)at^2}}}


{{{40.4 = (1/2)(9.81)t^2}}}


{{{t = (2.87) s}}}


The {{{total}}}{{{ time}}} is {{{(2.04)s + (2.87)s = (4.91) s}}}.