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

We are given that the velocity is proportional to certain power of the gravitational acceleration g and the distance h. We can insert the dimensional formula of the various quantities in the given formula for velocity and solve them for the values of unknown powers of h and g.

Complete step by step answer:
We are given an object which is falling freely under the influence of the gravitational force of earth. It is given that the velocity of the object after falling to distance h is equal to v. Now this velocity v depends on the acceleration due to gravity g and also on the distance h covered by it during its fall. Therefore, we can write that the velocity is proportional to h and v in the following way.
$v \propto {g^x}{h^y}$
To remove the proportionality, we can write that
$v = k{g^x}{h^y}$ …(i)
Here k is the constant of proportionality. Now we need to insert the dimensional formula of the various quantities on the either side of this equation. Doing so, we get
$\left[ {{M^0}{L^1}{T^{ - 1}}} \right] = {\left[ {{M^0}{L^1}{T^{ - 2}}} \right]^x}{\left[ {{M^0}{L^1}{T^0}} \right]^y} \\\ \left[ {{M^0}{L^1}{T^{ - 1}}} \right] = \left[ {{L^x}{T^{ - 2x}}} \right]\left[ {{L^y}} \right] \\\ \left[ {{M^0}{L^1}{T^{ - 1}}} \right] = \left[ {{L^{x + y}}{T^{ - 2x}}} \right] \\\$
Now we can equate the powers of the various quantities on either side. Doing so, we get
$x + y = 1 \\\ \- 2x = - 1 \\\$
The second equation gives us value of x, which is
$x = \dfrac{1}{2}$
Using this value of x in first equation, we get
$x + y = 1 \\\ \dfrac{1}{2} + y = 1 \\\ y = 1 - \dfrac{1}{2} \\\ y = \dfrac{1}{2} \\\$
Now we can insert these values of x and y in the equation (i). Doing so, we get
$v = k{g^{\dfrac{1}{2}}}{h^{\dfrac{1}{2}}} \\\ v = k\sqrt {gh} \\\$
This expression is the same as the given expression. Therefore, the given expression is correct.

Note:
There is another way to check if the given expression is dimensionally correct or not, by directly inserting the dimensions on the either side of the given expression. If the dimensions of the left hand side match the dimensions of the right hand side, then the given formula is dimensionally correct.