Question

An object is falling freely under the gravitational force. Its velocity after traveling a distance h is v. If v depends upon gravitational acceleration g and distance. Prove with dimensional analysis that \[v=k\sqrt{gh}\],where k is a constant.

Answer 1

Hint: This is a problem of dimensional analysis. we can deduce the value from the dimensional formula of these individual elements by finding the degree of dependence of a physical quantity on another. The principle of consistency of two expressions can be used to find the equation relating these two quantities

Complete step by step answer:We know velocity is given by the formula, \[v=\dfrac{d}{t}\], where d is the displacement and t is the time.
Dimensions of v is \[[L{{T}^{-1}}]\]
Acceleration is defined as the rate of change of velocity and is given by the formula, \[a=\dfrac{v}{t}\], where v is the velocity and t is the time. Dimensions of a is \[[L{{T}^{-2}}]\]
Now given in the question the equation is \[v=k\sqrt{gh}\], here g is acceleration due to gravity and h is the distance, equating the dimensions of the LHS and RHS, we get,
\[[L{{T}^{-1}}]\]=k \[{{[{{L}^{2}}{{T}^{-2}}]}^{\dfrac{1}{2}}}\]
\[\Rightarrow [L{{T}^{-1}}]=[L{{T}^{-1}}]\]
Thus, the equation is dimensionally stable. Thus, the given equation in the problem is correct

Note:This method of finding out the relationship between two quantities or checking out whether the given equation is correct or not has some limitations. The quantities which are dimensionless gets hollowed out during the use of this method, thus, this is not at all 100% fool proof method. Also, many times it becomes quite complex.