Answer:[tex]v_0=v\sqrt{\frac{g-vc}{g+vc}}[/tex]
Explanation:
Given
v=initial velocity
resisting acceleration =cv
also gravity is opposing the upward motion
Therefore distance traveled during upward motion
[tex]v^2_f-v^2=2as[/tex]
Where a=cv+g
[tex]0-v^2=2(cv+g)s[/tex]
[tex]s=\frac{v^2}{2(cv+g)}[/tex]
Now let v_0 be the velocity at the ground
[tex]v^2_0-0=2(g-vc)s[/tex]
substituting s value
[tex]v^2_0=v^2\frac{g-vc}{(cv+g)}[/tex]
[tex]v_0=v\sqrt{\frac{g-vc}{g+vc}}[/tex]