We need the acceleration of gravity, g, to answer these questions. Assume g=9.81m/s/s.
The equation is v^2=2gs where s is the distance fallen and v is the final speed.
Therefore v^2=2*9.81*80=1569.6, making v=39.62 m/s approx.
The equation s=ut-gt^2/2 where u is the vertical speed=50 m/s, t is time in seconds.
We can write this as 2s/g=2ut/g-t^2 or t^2-2ut/g+2s/g=0, a quadratic in t.
Complete the square:
t^2-2ut/g+u^2/g^2 - u^2/g^2+2s/g=0=(t-u/g)^2-u^2/g^2+2s/g.
The minimum value of (t-u/g)^2 is 0 and is achieved when t=u/g. This corresponds to the maximum height before the body starts to fall under gravity. This value of t means that -u^2/g^2+2s/g=0 and s=u^2/2g.
This evaluates to s=2500/19.62=127.42m.