The value of G can be measured very accurately by timing the vertical motion of ball as it rises past a pair of light gates separated by a known vertical distance-h.
If a ball spent a total time T1 above the lower light gate and a total time T2 above the upper light gate,show that g=8h / [ (T1)^2 - (T2)^2 ].
Because of the symmetry of motion in the problem, the time taken to go from 1 to 2 is,
T1/2 - T2/2
hence, where u is the velocity at 1
h = u(T1/2 - T2/2) - 1/2 g (T1/2 - T2/2)^2
at the maximum height v=0 => u=g(T1/2)
Thus you will get,
g = 8h / [ (T1)^2 - (T2)^2 ]
ALTERNATIVELY,
you can use the work-energy principle, and since the mass is constant the equation is reduced to
1/2 [ g (T2/2) ]^2 - 1/2 [ g (T1/2) ]^2 = -gh
then solve for g in terms of h and T1 and T2 you get,
g = 8h / [ (T1)^2 - (T2)^2 ]
which is exactly the same thing!
DIAGRAM
------- max height , v=0 => u=g(T1/2)
------- sensor 2 , v= g (T2/2)
-------- sensor 1 , v=u
-------- origin / "ground" v=unknown, but no need to care
well done audioboxing! Only thing is the way of presentation needs to be a bit clearer... I got confused a little initially by your solution until I went to do it myself...
The confusing line is "at the maximum height, u=g(T1/2)"
u is a constant in its usage above, so there's no need to state "at max height". Instead, we should phrase it like
v = u - g(T1/2)
At max height, v = 0
Therefore, u=g(T1/2)
I think it would be clearer this way, to the readers, to markers, and most importantly to the students seeking an understandable solution.
Thanks! :D
Ah...yes I made a mistake in the phrasing over there...initially wanted to type at max height v is 0 hence u= ...(mind move faster than fingers on the keyboard, lol)
Sorry if I confused anyone! :D