A falling flower pot (kinematics question), just need slight correction

[Sean1218]

Homework Statement

A falling flowerpot takes 0.2s to fall past a window that is 1.9m tall. From what height above the top of the window was the flower pot dropped? g = -9.8 m/s^2, v1 = 0 m/s

Homework Equations

d = v1t + 1/2at^2
a = (v2 - v1) / t
d = (v2 + v1) / 2 * t

The Attempt at a Solution

Ok, so, I sort of got the right answer, but I'm doing something wrong with my negative signs.

I plugged the values into the first equation to solve for v2 (speed at the bottom of the window), then plugged the values into the second equation (this time with v1 as 0 m/s (speed at where flower pot is dropped i.e. rest)), and got a negative value.

(8.52 - 0) / -9.8 = t = - 0.87

Then I plugged that time value into the 3rd equation to solve for d, which would give me a negative height value (-3.7, while the answer is 3.7). So, where did I mess up with my negatives?

 

[rl.bhat]

If you have taken g negative then d and v must be negative.

Try this approach.

Let d be the distance from the starting point to the top of the window.

Then the distance between the starting point and the bottom of the window is ( d + 1.9 m). If t is the time taken by the flowerpot to reach the top of the window, then

d = 1/2*g*t^2 ...(1)

(d + 1.9 ) = 1/2*g*(t + 0.2 s)^2...(2)

Solve these two equations to find d and t.