Q. Determine the elapsed time in kinematics?
Let
S = 1 + c2+ c4+ ........
Then,
S = 1 + c2(1 + c2+ c4+.......... )
S = 1 + c2S so,
S - c2S = 1
S(1 - c2) = 1 and finally,
S =1/1 - c2
We now have,
d = h0 + 2c2h0 S
d = h0 +2c2h0/1 - c2
Given the initial height, h0, the distance can now be found. For example, if h0 = 6ft, d = 57.14 ft.
We at this time determined the elapsed time. From the kinematics equation,
v = g t, for an object moving under the influence of gravity we have,
c =v1/v0=t1/t0,
Where t0 is the time to fall from the initial height to the surface and t1 is the time to reach the peak of the first bounce. So t1 = c t0.
Then
t = t0 + 2t1 + 2t2 + 2t3 +
t = t0 + 2c t0 + 2c2t0 + 2c3t0 +
t = t0 + 2t0(c + c2+ c3+ )
Now let
S = c + c2+ c3+
S = c + c(c + c2+ c3+ ) Then
S = c + c S so,
S - c S = c and
S =c/1 - c.
We have
t = t0 + 2t0 S = t0 +2c t0/1 - c.
From the kinematics equation, h = ½*g t2
t0 =√2h0/g.
Finally,
t = t0 (1+(2c/(1 - c))=√2h0/g(1 +(2c/(1 - c))
From the preceding problem with initial height, h0 = 6ft,
t = 11.64sec.