The acceleration due to gravity at two places are g and `g^'` respectively. A body is dropped from the same height at both places. At the second place, the required time to touch the ground is t seconds less than that in the first place, while the velocity attained in reaching the ground is higher by a value of v than that in the first place. Show that `gg^'=(v^2)/(t^2)`.

Explanation

Let the time taken by the body to fall at the first place be T and velocity with which the body touches the ground be V.
Hence, the time taken to reach the ground and the corresponding velocity are (T-t) and V+v in the second place .
Let the body fall from a height h in each case. Considering the motion at the first place,
`h=1/2 gT^2 or, T=sqrt((2h)/g)" ".......(1)`
and `V^2=2gh or, V=sqrt(2gh)" " ........(2) `
Similarly , for motion at the second place,
`h=1/2 g^.(T-t)^2 or, T-t =sqrt((2h)/(g^.))" "......(3) `
and `(V+v)^2=2g^.h or, V+v=sqrt(2g^.h)" ".......(4)`
Subtracting (3) from (1)
`t=sqrt((2h)/g)-sqrt((2h)/(g^.))=sqrt(2h)(1/sqrt(g)-1/sqrt(g^.))`
Subtracting (2) from (4)
`v=sqrt(2g^.h)-sqrt(2gh)=sqrt(2h)(sqrt(g^.)-sqrt(g))`
`therefore (v^2)/(t^2)=(2h(sqrt(g^.)-sqrt(g))^2)/(2h((sqrt(g^.)-sqrt(g))/sqrt(gg^.)))=gg^.`.