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Old Sep 28th 2017, 03:29 PM   #1
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Calculating g with time intervals of thrown object.

At the National Physics Laboratory in England a measurement of the gravitational acceleration g was made by throwing a glass ball straight up in an evacuated tube and letting it return, as shown in the Figure. The time interval between the two passages across the lower level is equal to ∆TL = 2.70 s. The time interval between the two passages across the upper level is equal to ∆TH = 1.03 s. The distance between the two levels is equal to H = 7.69 m. Calculate the magnitude of g. Note that the unknown in this question is g, and that this is one-dimensional motion.

https://imgur.com/a/tf4jI

Can someone please solve this for me and explain the steps, ive been going at it for hours and have gotten absolutely nowhere.
Thanks
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Old Sep 28th 2017, 06:12 PM   #2
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$\Delta y = v_{y0}t - \dfrac{1}{2}gt^2$

for both given time periods $\Delta y =0 \implies t\left(v_{y0}-\dfrac{1}{2}gt\right)=0 \implies v_{y0}=\dfrac{1}{2}gt$

Using the equation $v_f^2=v_0^2 -2g \Delta y$ on the projectile’s way upward ...

$v_{y0H}^2 = v_{y0L}^2 -2gH$

$\dfrac{1}{4}g^2(\Delta T_H)^2=\dfrac{1}{4}g^2(\Delta T_L)^2 -2gH$

$g = \dfrac{8H}{(\Delta T_L)^2 - (\Delta T_H)^2}$
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