A pellet gun is fired straight downward from the edge of a cliff that is 12 m ab

Question

A pellet gun is fired straight downward from the edge of a cliff that is 12 m above the ground. The pellet strikes the ground with a speed of 27 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

So here is what I did: V(f)^2= V(o)^2 + 2ax

V(f)= 27 m/s
x=12 m
a=gravity=-9.81 m/s^2


Plug it in and solve for V(o)
V(f)^2= V(o)^2 + 2ax
(27)^2 = V(o)^2 + 2(-9.81)(12)


I am being told that the gravity should be POSITIVE NOT NEGATIVE. I do not understand why because in all the other problems, I did gravity going down is negative and going up is positive. Any ideas?

Explanation / Answer

Use the given information for when the gun was fired upward to find the pellet’s initial velocity (v0). The direction is downward, so the velocity, displacement, and acceleration are all negative: v² = v0² + 2g?y v0 = v[v² - 2g?y] = v[(-27m/s)² - 2(-9.8m/s²)(-12m)] = -22.2m/s------------->select the minus root, it is downward. Now use the same formula (solved for ?y) to find the height it reaches. Since the gun is fired upward, the velocity is now upward, so it is positive. Acceleration remains negative as it is downward: ?y = (v² - v0²) / 2g = [0 - (22.2m/s)²] / 2(-9.8m/s²) = 24.89m

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