Why Does Freefall Acceleration Sign Differ Between Similar Problems?

[willcrys84]

I worked two similar freefall problems with the same coordinate system, but had different signs for acceleration on each solution. I thought that if we say +y is up, then a = -9.8 m/s², so I'm confused as to why a = +9.8 m/s² for the second problem.

Problem 1: Rock thrown from 30 m cliff with initial velocity of 12 m/s. Find v when it hits ground.
Coordinate system: +y is up and 0 is on ground. (so a = -9.8 m/s²)
Solution: v_f² = v₀² + 2a(x - x₀) = (12 m/s)² + 2(-9.8 m/s²)(0 m - 30 m) = 732
v_f = 27.1 m/s
This makes sense to me because if a = +9.8 m/s², I'd have to take the square root of a negative.

Problem 2: A rock passes a 3m window in 0.4 s. At what height above the window was it dropped from.
Coordinate system: +y is up, zero at bottom of window. (so a = -9.8 m/s²)
Solution: Let d be the distance from the drop point to the top of the window. First find velocity at top of window. Assume rock dropped from rest, then find d.
(x - x₀) = v₀t + ½at²
(0 m - 3 m) = 0.4v₀ + ½(-9.8 m/s²)(0.4 s)²
v₀ = -5.54 m/s
v_f² = v₀² + 2a(x - x₀)
(-5.54 m/s)² = 0 + 2(-9.8 m/s²)[(d + 3) m - 3 m]
d = -1.57 m
Which doesn't make sense physically, because it would be below the bottom of the window.So I tried it with a = +9.8 m/s², I get -9.46 m/s for velocity at the top of the window and d = 4.56 m.

 

[tiny-tim]

welcome to pf!

hi willcrys84! welcome to pf!

Problem 1: Rock thrown from 30 m cliff with initial velocity of 12 m/s. Find v when it hits ground.

it doesn't make any difference whether it's 12 m/s up or 12 m/s down, does it?

v_f² = v₀² + 2a(x - x₀)

x₀ is the top, so (x - x₀) is negative