Calculus-based physics mechanics

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The discussion revolves around a physics problem involving a glass ball projected in an evacuated tube, requiring the derivation of an expression for gravitational acceleration, g, in terms of the distance H and the time intervals TsubL and TsubU. Participants clarify that the time interval between upward and downward crossings can be expressed as Δt = 2v/g, where v is the velocity at a given height. The equations derived from the problem involve manipulating the initial velocity and gravitational effects, leading to a system of equations to solve for g. There is a debate about the appropriateness of using g in the equations, with clarification that it should be treated as an unknown to be solved. The conversation emphasizes the importance of correctly interpreting the problem requirements to derive the desired expression for g.
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Calculus-based physics...mechanics

I'm having a little trouble with this problem...

"A small glass ball is projected straight up in an evacuated tube (no air resistance) and then falls back down. During its motion it passed both an upper and a lower timer that are separted by a distance H.

Let TsubL be the time interval between the two passes across the lower timer, and TsubU be the time interval between the two passages across the upper timer.

Find an expression for the acceleration of gravity, g, in terms of H, TsubL, and TsubU.

Any help would be appreciated, but thorough explanation would be a plus. Thanks everybody.
 
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For any given vertical location the time interval between (upward and downward) crossings is \Delta t = \frac {2v}{g}. If the initial speed is v_0 then, for a given height, v = \sqrt {v_0^2 - 2gy} and using the information provided we have
g^2 T_L^2 = 4 \left( v_0^2 - 2g y_L \right)
g^2 T_U^2 = 4 \left( v_0^2 - 2g y_U \right)
Now subtract the equations and solve for g!
 
Thanks a lot Tide, but I don't think I am allowed to use acceleration due to gravity in the equations, because i am trying to define g. It's like using a word in its own definition. I don't think my teacher will let it fly.
 
I don't think that is what your teacher means. He or she is asking you to calculate the value of g which means it becomes an unknown value in your equations and you must solve for it. That is what is meant by "find an expression for."
 
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