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

z_surface = 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)

v_surface = 10 + 1.6(7.5) = 22 s

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of course using your equation

(v_surface)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.