A particle is thrown vertically upwards from the surface of the earth. Let Tp be the time taken by the particle to travel from a point P above the earth to its hightst point and back to P. Similarly let Tq be the time taken by the particle to travel from a point Q above the earth to its hightst point and back to Q. If the distance between the points P and Q is H then find the expression of acceleration due to gravity in terms of Tp, Tq and H

Dear Student,
according to the question

TQ is the time taken by the particle to travel from another point Q above the earth to its highest point and back to the same in terms of TP, TQ and H, is distance between P and Q

Using equations of motion,

h = vi × t + 0.5 × g × t²

Since the object is falling, vi = 0.

rearranging the terms, t = √(2h/g)

So, we have obtained the equation forfalling time,

T = √(2h/g)

Using the attached diagram, we have,

Time taken from point P to point P = TP = 2√[2(h + H)/g]

⇒ TP² = 8(h + H)/g

∴TP² = 8h/g + 8H/g.......(1)

Time taken from point Q to point Q = TQ = 2√[2h/g]

⇒TQ² = 8h/g.......(2)

comparing equations (1) and (2), we get,

TP² = TQ² + 8H/g

8H/g = TP² - TQ²

g = 8H/[TP² - TQ²]

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Yuvraj Singh