An object is falling freely under the gravitational force. Its velocity after travelling a distance his v. If v depends on gravitational accelertation g and distance h, then with the help of dimensional analysis, formula of v is ....... (k is constant)

Explanation

Let `v prop g^(a)+h^(b)` where `a,b , in R`,
`[v]=L^(1)T^(-1),[g]=L^(1)T^(-2),[h]=L^(1)`
`:. V= kg^(a)h^(b)=...(i)` (k is constant)
`:.[v]=k[g]^(a) [h]^(b)`
`:. L^(1)T^(-1)=(L^(1)T^(-2))^(a) (L^(1))^(b)`
`:. L^(1)T^(-1)=L^(a+b)T^(-2a)`
Comparing powers of L, we get `a,+b=1... (ii)`
Comparing powers of T. we get
`-2a=1`
`:. a=(1)/(2)`
Putting this value eq. (ii) `(1)/(2)+b=1`
`:.b=(1)/(2)`
`:.` In eq (i). If we put `k=1, a=(1)/(2) and b=(1)/(2)`, we get `v=kg^((1)/(2))h^((1)/(2))=k sqrt(gh)`