A spaceship decents at a constant velocity of 10m/s on the moon. When it is 120m from the moon's surface, an object falls off the spaceship. Acceleration due to gravity is 1.6m/s^-2, What is the velocity of the object when it reaches the surface of the moon

using
V^2=U^2-2AS
with downwards as positive , I get a different answer from the book's answer which is 22m/s
and can be obtained with V^2=U^2+2AS (WHYY?? I've taken note of the vectors) and I still get 16.8m/s

ARGHHHH!! PLEASE PLEASE HELP
Got a test coming soon!

:o

z = z0 + v0*t + 1/2*a*t2

dz/dt = v = v0 + a*t

v0 = 10 m/s

a = 1.6 m/s2

z0 = 0

zsurface = 120 m

120 = 10*t + 1/2*1.6*t2

0 = -120 + 10*t + 0.8*t2 (quadratic equation)

a = 0.8, b = 10, c = -120

t = (-b +/- [b2 - 4*a*c]1/2)/2*a

t = (-10 +/- [102 - 4*0.8*(-120)]1/2)/2*(0.8)

t = (7.5, -20) (can't have negative time so t = 7.5s)

vsurface = 10 + 1.6(7.5) = 22 s

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of course using your equation

(vsurface)2 = (v0)2 + 2*a*z

v0 = 10 m/s

a = 1.6 m/s2

z = 120m

v_surface = (100 + 2*1.6*120)1/2 = 22 s

if down is positive, that means your at 120 m when you hit the surface. That is why down should be negative so you are at zero when you hit the surface.