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.