- #1

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So we have time and dinstance, but not the acceleration, which confuses me terribly. Do I need to find velocity? Or do I need to look for the acceleration? SO confused. Any help will be greatly appreciated.

Thanks!!

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- Thread starter TickleMeElma
- Start date

- #1

- 16

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So we have time and dinstance, but not the acceleration, which confuses me terribly. Do I need to find velocity? Or do I need to look for the acceleration? SO confused. Any help will be greatly appreciated.

Thanks!!

- #2

hotvette

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- #4

hotvette

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Let me try this way. You have an equation of motion that relates distance fallen (from an unknown point above the window) as a function of time. Seems to me a fruitful approach is to express that equation of motion for [itex]x_1, t_1[/itex] and [itex]x_2, t_2[/itex], where [itex]x_1[/itex] represents the top of the window and [itex]x_2[/itex] represents the bottom of the window. Subtracting the 2 equations gives you [itex]x_1 - x_2[/itex], which you know.

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- #6

hotvette

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Had to do a little head scratching. I believe the following will work:

Using the equation of motion of the rock you should be able to get an equation for [itex]x_1-x_2[/itex] in terms of [itex]g[/itex], [itex]t_1^2[/itex] and [itex]t_2^2[/itex]. You know [itex]x_1-x_2[/itex], and [itex]g[/itex], but not [itex]t_1[/itex] or [itex]t_2[/itex]. But, you do know [itex]t1-t2[/itex]. Thus, you have two equations in two unkowns and should be able to solve for [itex]t_1[/itex], back substitute into the equation of motion, and find [itex]x_1[/itex].

Using the equation of motion of the rock you should be able to get an equation for [itex]x_1-x_2[/itex] in terms of [itex]g[/itex], [itex]t_1^2[/itex] and [itex]t_2^2[/itex]. You know [itex]x_1-x_2[/itex], and [itex]g[/itex], but not [itex]t_1[/itex] or [itex]t_2[/itex]. But, you do know [itex]t1-t2[/itex]. Thus, you have two equations in two unkowns and should be able to solve for [itex]t_1[/itex], back substitute into the equation of motion, and find [itex]x_1[/itex].

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- #7

hotvette

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