- #1

spj1

- 7

- 0

**1. A flower pot falls out of a window and past the window below (ignore air resistance). It takes the post .42 s to pass this window, which is 1.9m high. How far is the top of the window below the windowsill from which the flowerpot fell?**

answer: .31 m

answer: .31 m

**2.**

v = v0 + at

x = x0 + v0t + 1/2at^2

v^2 = v0^2 + 2a(x - x0)

v = v0 + at

x = x0 + v0t + 1/2at^2

v^2 = v0^2 + 2a(x - x0)

**3. The Attempt at a Solution**

Known facts:

t = .42 s

bottom window height = 1.8

a = -g

x0 = 0 (if the origin is the top windowsill)

v0 = ?

x = ?

x = 0 + v0(.42) + 1/2(-9.8)(.42)^2

So I need to find the initial velocity so I can get the total distance and subtract it from the 1.9. If I had the total distance down, then I could get it easy, but I'm stumped from here since I'm trying to find the total distance. I've tried using v0=0 and that didn't work. I also tried setting the v = 0 (thinking I could do that sense the problem ends with that bottom window). Every route I go it seems I'm still left with two unknowns or I end with a total distance of .86 which isn't even longer than the bottom window.

Known facts:

t = .42 s

bottom window height = 1.8

a = -g

x0 = 0 (if the origin is the top windowsill)

v0 = ?

x = ?

x = 0 + v0(.42) + 1/2(-9.8)(.42)^2

So I need to find the initial velocity so I can get the total distance and subtract it from the 1.9. If I had the total distance down, then I could get it easy, but I'm stumped from here since I'm trying to find the total distance. I've tried using v0=0 and that didn't work. I also tried setting the v = 0 (thinking I could do that sense the problem ends with that bottom window). Every route I go it seems I'm still left with two unknowns or I end with a total distance of .86 which isn't even longer than the bottom window.

Any hints on which equations to put together?