Help me with this problem Pleaseeee!?

Question

A parachutist descending at a speed of 9.40 m/s loses a shoe at an altitude of 31.60 m.

(b) What is the velocity of the shoe just before it hits the ground?

Answer 1

♦ final speed v=u+g*t, where u=9.4 m/s,
hence time of free fall t=(v-u)/g;
♥ distance of free fall s=u*t +0.5*g*t^2 =
= u*(v-u)/g +0.5*(v-u)^2/g; or;
(v-u)^2 +2u(v-u) –2gs=0, hence v-u = -u +√(u^2 +2gs);
♣ v=√(u^2 +2gs) =√(9.4^2 +2*9.8*31.6) =26.60 m/s;

Answer 2

s = s(zero) + v(zero)t + (1/2)at^2
s = 31.60 + 9.4t + (1/2)gt^2
s = 31.60 + 9.4t - 4.9t^2
4.9t^2 - 9.4t - 31.60 = 0
t = (-B =/- √[B^2 - 4AC])/2A
t = (9.4 +/- √[(-9.4)^2 - 4(4.9)(-31.60)]/9.8
t = (9.4 +/- 26.603).9.8
The positive answer makes sense.
t = 3.6738s
v = v(zero) + at
v = 9.4 + (9.8)(3.6738) = 45.40m/s