- #1
ProBasket
- 140
- 0
As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time [tex]t[/tex] , and the vertical length of your window is [tex]L_w[/tex]. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity [tex]g[/tex] .
Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).
From what height [tex]h[/tex] above the bottom of your window was the flower pot dropped?
Express your answer in terms of [tex]L_W, t,[/tex] and [tex]g[/tex].
I'm trying to find the position right? so I'm going to use the position formula:
x = x(0) + v(0)t + 1/2at^2
well i know that initial velocity is 0, so...
x = [tex]L_W + 1/2*g*t^2[/tex]
is this correct?
Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).
From what height [tex]h[/tex] above the bottom of your window was the flower pot dropped?
Express your answer in terms of [tex]L_W, t,[/tex] and [tex]g[/tex].
I'm trying to find the position right? so I'm going to use the position formula:
x = x(0) + v(0)t + 1/2at^2
well i know that initial velocity is 0, so...
x = [tex]L_W + 1/2*g*t^2[/tex]
is this correct?