Calculating Height of Dropped Water Balloons Above Window Edge

[Zeye1280]

As you look out of your apartment window, you see water balloons falling past. You measure that the balloons are visible (going past the window)for a time t, and the vertical length of your window is L_w. assume that the water balloons were dropped (rather than thrown downward). Determine then height h (above the top edge of your window) from which it was dropped. This will be an algebraic expression in terms of t, L_w, and g.

my approach:
known's:
a_y= -g
Vi_y= 0
Vf_h=Vi _(L_w)

Picture I made to help with solution
http://yfrog.com/5hbaloonfallingj

My use of kinematics
Vf _h= -g*t which also means Vi_(L_w) = -g*t
Yf_h = Yi_(L_w)
L_w = Yi_(L_w) - g*t*t - (1/2)g*t^2 ------> Yi_(L_w) = L_w +(1/2)g*t^2
Yf_h = Yi_h - (1/2)g*t^2 ----> L_w +(1/2)g*t^2 = Yi_h -(1/2)g*t^2 --(solve for Yi_h)--> Yi_h = g*t^2 + L_w
Answer: [Yi_h = g*t^2 + L_w] is the algebraic expression to find height above the top of window edge.

This is what I came up with and wondering if this is the correct answer for this problem?

 

[ehild]

No, it is not correct. Try to write your derivation a bit more clearly. First, choose your reference system with the y-axis pointing downward and the origin at the point from where the balloon was dropped. The y coordinate of the top of the window is y₁=h. The bottom is at y₂=h+L.

At time t₁ the balloon is at

$$h=\frac{1}{2}t_1^2$$

and at time t₂ it is at


$$h+L=\frac{1}{2}t_2^2$$


ehild

You measure the time span while the balloon moves from h to h+L, this is t₂-t₁=t. So t₂=t₁+t.

Insert this for t₂ in the second equation, determine t₁, plug into the first equation: you get h.