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.(adsbygoogle = window.adsbygoogle || []).push({});

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_{f}² = v_{0}² + 2a(x - x_{0}) = (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 - x_{0}) = v_{0}t + ½at²

(0 m - 3 m) = 0.4v_{0}+ ½(-9.8 m/s²)(0.4 s)²

v_{0}= -5.54 m/s

v_{f}² = v_{0}² + 2a(x - x_{0})

(-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.

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# Freefall acceleration

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