Question

Example 3.43 A particle is thrown vertically upwards from
the surface of the earth. Let To be the time taken by the
particle to travel from a point P above the earth to its highest
point and back to the point P.
Similarly, let To be the time taken by the particle to travel from
another point ( above the earth to its highest point and back to
ihe same point Q. If the distance between the points P and Q is
H, find the expression for acceleration due to gravity in terms
of Tp, TQ and H.​

Answer 1

Explanation:

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Answer 2

From point P to point O it will take half the time as it takes while going from point P to point P.

$\dfrac{{T}_{P}}{2} = \sqrt{\dfrac{2(H+x)}{g}}$

Squaring both sides.

$\dfrac{{T}_{P}^{2} }{4} = \dfrac{2(H+x)}{g}$

Now, from point Q to point O it will take same as in above condition.

$\dfrac{{T}_{Q}}{2} = \sqrt{\dfrac{2x}{g}}$

In this also squaring both sides.

$\dfrac{{T}^{2}_{Q}}{4} = \dfrac{2x}{g}----(2)$

$\dfrac{{T}^{2}_{Q}}{4} - \dfrac{{T}^{2}_{Q}}{4} = \dfrac{2H+2x}{g} - \dfrac{2x}{g}$

$\dfrac{{T}^{2}_{P}-{T}^{2}_{Q}}{4} = \dfrac{2H}{g}$

$g = \dfrac{8H}{{T}^{2}_{P}-{T}^{2}_{Q}}$