We will have the following:
a) The velocity after 5 seconds will be:
[tex]v_f=v_o+at[/tex][tex]v=2.79m/s+(9.8m/s^2)(5s)\Rightarrow v=51.79m/s[/tex]So, the velocity after 5 seconds will be 51.79m/s.
b) We will have that the distance below the helicopter will be:
[tex]d=\frac{1}{2}(v_0+v_f)t[/tex][tex]d=\frac{1}{2}(2.79m/s+51.79m/s)(5s)\Rightarrow d=136.45m[/tex]So, it will be 136.45 m below the helicopter.
c) We will have that the velocity and distance given that the helicopter is moving constantly at 2.79m/s will be:
[tex]v=-2.79m/s+(9.8m/s^2)(5s)^2\Rightarrow v=46.21m/s[/tex]So, the velocity would be 46.21 m/s.
[tex]d=\frac{1}{2}(-2.79m/s+46.21m/s)(5s)\Rightarrow d=108.55m[/tex]So, the distance would be 108.55 m below the helicopter.