JC Kinematics

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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