Determine the elapsed time in kinematics, Physics

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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.


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