Solving Kinematics: Finding Release Height Above Window

[rjs123]

Homework Statement

A water balloon takes .22 s to cross the 130 cm high window, from what height above the top of the window was it dropped?

The Attempt at a Solution

I'm using:

v^2 = vo^2 + 2a(x - xo)

v = distance/time
v = 1.30 m / .22 s = 5.91 m/s

(5.91 m/s)^2 = 0 + 2(9.8 m/s^2)(x)
x = 1.78 m from release to bottom of window

1.78 m - 1.3 m = distance of release from top of window

.48 m


I'm doing something wrong since the available answers aren't .48m...and it is something simple.

 

[Pi-Bond]

You can't simply use distance divided by time to get the velocity at the bottom of the window, as the balloon is being accelerated due to the earth. What other equation could you apply?

 

[rjs123]

You can't simply use distance divided by time to get the velocity at the bottom of the window, as the balloon is being accelerated due to the earth. What other equation could you apply?

nevermind i got it.

 

[Pi-Bond]

Right, you can solve for v₀ from that equation, and use that to obtain the distance above the top of the window the ball was thrown from.

 

[rjs123]

Right, you can solve for v₀ from that equation, and use that to obtain the distance above the top of the window the ball was thrown from.

yeah i see what i did...thanks.