Question

The speed v of an object falling with a constant acceleration g can be expressed in terms of g and the distance traveled from the point of release, h, as v = kgp hq, where k, p, and q, are dimensionless constants. What must be the values of p and q?

Answer 1

So the values are
`p= (1)/(2)` ,
`q= (1)/(2)`

Step-by-step explanation:

From the question we are told that

The equation is
`v = k [g^p][h^q]`

Now dimension of v (speed ) is

`v = m/s = LT^(-1)`

Now dimension of g (acceleration ) is

`g= m/s^2 = LT^(-2)`

Now dimension of h (vertical distance ) is

`h= m = L`

So

`LT^(-1) = [ [LT^(-2)]^p][[ L]^q]`

`LT^(-1) = [ [T^(-2p)][[ L]^(p +q)]`

Equating powers

`1 =p+q`

`-1 = -2p`

=>
`p= (1)/(2)`

and

`q= 1 -(1)/(2) = (1)/(2)`