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A particle is thrown vertically upwards ...

A particle is thrown vertically upwards from the surface of the earth. Let `T_(P)` 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 `T_(Q)` be the time taken by the particle to the particle to travel from another point Q above the earth to its highest point and back to the same point Q. if the distance between the point P and Q is H, find the expression for acceleration due to gravity in terms of `T_(P).T_(Q)` and H.

A

`(6H)/(T_(P)^(2)+T_(Q)^(2))`

B

`(8H)/(T_(P)^(2)+T_(Q)^(2))`

C

`(2H)/(T_(P)^(2)+T_(Q)^(2))`

D

`(H)/(T_(P)^(2)+T_(Q)^(2))`

Text Solution

Verified by Experts

The correct Answer is:
B

Time taken from from point P to point P `" " T_(P)=2sqrt((2(h+H))/g)`
Time taken from point Q to point Q `" "T_(Q)=2sqrt((2h)/g)`
`rArr T_(P)^(2)=(8(h+H))/g` and `T_(Q)^(2)=(8h)/g rArr T_(P)^(2)=T_(Q)^(2)+(8H)/g rArr g=(8H)/(T_(P)^(2)-T_(Q)^(2))`
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