Measuring acceleration of gravity

[tronter]

[SOLVED] Measuring acceleration of gravity

Homework Statement

The acceleration of gravity can be measured by projecting a body upward and measuring the time that it takes to pass two given points in both directions.

Show that if the time the body takes to pass the horizontal line $$A$$ in both directions is $$T_A$$, and the time to go by a second line $$B$$ in both directions $$T_B$$, then, assuming that the acceleration is constant, its magnitude is $$g = \frac{8h}{T_{A}^{2} - T_{B}^{2}}$$ where $$h$$ is the height of line $$B$$ above line $$A$$.

Homework Equations

The Attempt at a Solution

I am not sure how to approach this. I know that $$g = 9.8$$. The path the body takes is a parabola. And $$a = \dot{v}$$.

 

[tronter]

Since the path is a parabola, I would expect there to be a quadratic term?

 

[robphy]

Draw a y-vs-t graph of the motion... a parabola.
Mark the two heights with $$y_B> y_A$$.
Note that there are four events.. label them sequentially as "1" (at A), "2" (at B), "3" (at B), "4" (at A).
Do you know any relationships among any of the quantities at those events?

 

[robphy]

By the way, this sounds like a Kleppner-Kolenkow problem.

 

[tronter]

yes it is. I already solved it.

Thanks

 

[robphy]

Great!
Did my hint help? Or did you come up with it yourself?
Or did you use a different approach?

[When solved, you can use the Thread Tools menu above to "Mark this thread as Solved".]

 

[tronter]

Yeah I just used $$y = y_0 + v_{y0}t - \frac{1}{2}gt^{2}$$ where we consider $$y_0 = y_A$$ and $$y_0 = y_B$$.

Then solve a quadratic, subtract, and rearrange.

 

[robphy]

I see. Good.
My method avoids solving a quadratic explicitly by using
the velocity and velocity-squared kinematic equations and some symmetry.

 

[John O' Meara]

I have it now as I found the same problem in classical physics Thanks anyway