# 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 r

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?

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1 year 2021-08-23T19:57:14+00:00 1 Answers 23 views 0

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

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