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

Answer:

So the values are  $p= \frac{1}{2}$ , $q= \frac{1}{2}$

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

       $1 =p+q$

       $-1 = -2p$

=>      $p= \frac{1}{2}$

and

        $q= 1 -\frac{1}{2} = \frac{1}{2}$