Solving Kinematics: Finding Release Height Above Window

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Homework Help Overview

The problem involves a kinematics scenario where a water balloon is dropped, and the goal is to determine the height from which it was released above a window that is 130 cm high. The original poster attempts to calculate the release height using kinematic equations but expresses uncertainty about their result.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find the initial velocity and height, with some questioning the validity of calculating velocity by simply dividing distance by time due to the influence of acceleration from gravity.

Discussion Status

Some participants have provided guidance on alternative approaches to find the initial velocity, suggesting that the original poster reconsider their method. There appears to be a productive exchange of ideas, although no consensus has been reached on the correct solution.

Contextual Notes

The original poster indicates that their calculated height does not match the available answers, suggesting potential constraints or misunderstandings in the problem setup or assumptions made during calculations.

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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.
 
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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?
 
Pi-Bond said:
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.
 
Last edited:
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.
 
Pi-Bond said:
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.
 

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