Question

An object is falling freely under the gravitational force. Its velocity after travelling 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: In order to solve this question, the concept of dimensional analysis is important. First, analyse which quantities are given and what are their respective dimensions. After getting the dimensions of the quantities known, try to establish the relationship between them.

Complete step by step solution:
First of all, Let us consider a point A where the initial velocity is zero.
After some time, after covering a distance $h$ we have the final velocity is equal to $v$ .
Now, we have to find the relationship between $v$ , $g$ and $h$ .
We have to use the concept of dimensional analysis, so let us consider,
${v^x} = {g^y}{h^z}$
As we know that the dimensions of velocity are given by meter/second hence we have,
$v = \dfrac{{[L]}}{{[T]}}$
This can also be written as,
$v = [L{T^{ - 1}}]$
Now, the dimensions of \[g\] is the same as that of acceleration because it is the acceleration due to gravity. As the dimensions of acceleration is meter per second’s square. So, we have,
$g = \dfrac{{[L]}}{{[{T^2}]}}$
This can also be written as,
$g = [L{T^{ - 2}}]$
Now, the dimension of height is given in meters. So, we have,
$h = [L]$
Now, putting the calculated expressions of dimensions in ${v^x} = {g^y}{h^z}$ we have,
${[L{T^{ - 1}}]^x} = {[L{T^{ - 2}}]^y}{[{L^{}}]^z}$
Using the rule of multiplication of powers we have,
${L^x}{T^{ - x}} = {L^{y + z}}{T^{ - 2y}}$
On comparing both the sides we have,
$x = y + z$
And also, $x = 2y$
On comparing the above two equations we have $x$ on both the sides which deduce,
$y + z = 2y$
Or we can say, $y = z$
Putting $y = z$ and $x = 2y$ in equation ${v^x} = {g^y}{h^z}$ we have,
${v^{2y}} = {g^y}{h^y}$
As $y$ is common on both the sides of the powers hence we can cancel it, so we have now,
${v^2} = gh$
This can be written as,
$v = k\sqrt {gh} $
Hence proved.

Note: There are many useful applications of dimensional analysis. One application is to check the consistency of a given equation, another application is to derive the relationship between the given physical quantities. We have used the later one to solve this question.