Question

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

Answer 1

Use the third question of motion which states that${v^2} - {u^2} = 2as$. The dimension for velocity is $L{T^{ - 1}}$ and for acceleration is $L{T^{ - 2}}$.

Complete step by step answer:
An object is falling freely under the gravitational force. Its velocity after traveling a distance $h$ is $v$.
Since the object is falling freely, therefore, the initial velocity of the object will be zero.
Velocity of the object after traveling a distance $h$ is $v$.
Therefore, the distance travelled which is $S$ is $h$
Final Velocity is $v$
And acceleration due to gravity is $g$
Therefore,
$a = g$
Now, we have to find relation between $v,g$ and $h$
Let us assume that,
${v^x} = {g^y} {h^z}$
Now, we know that velocity is ratio of distance upon time
$v = \dfrac{m}{s}$ or
$v = \dfrac{{[L]}}{{[T]}}$
$v = [L {T^ { - 1}}]$
Also,
$g = \dfrac{m}{{{s^2}}}$
$g = \dfrac{{[L]}}{{{{[T]}^2}}}$
$g = [L {T^ {- 2}}]$
And height
$h = [L]$
Therefore, putting the values in equation
${v^x} = {g^y} {h^z}$
${[L{T^{ - 1}}]^x} = {[L{T^{ - 2}}]^y}{[L]^z}$
${L^x} {T^ {- x}} = [{L^y} {T^ {- 2y}}] [{L^z}]$
Comparing the powers,
We get
$x = y + z$
And
$- x = - 2y$
Therefore, $x = 2y$and $y = z$
Therefore, putting these values in the equation
${v^x} = {g^y} {h^z}$ ${v^2} = 2gh$
$v = \sqrt {2gh}$
Considering $\sqrt 2 = k$ where $k$ is the constant,
$v = k\sqrt {gh}$
Hence proved.

Note:
Alternative solution-
An object is falling freely under the gravitational force. Its velocity after traveling a distance $h$ is $v$.
Since the object is falling freely, therefore, the initial velocity of the object will be zero.
Velocity of the object after traveling a distance $h$ is $v$.
Therefore, the distance travelled which is $S$is $h$
Final Velocity is $v$
And acceleration due to gravity is $g$
Therefore,
$a = g$
Using the third equation of motion,
${v^2} - {u^2} = 2as$
Putting the given values,
${v^2} - 0 = 2gh$
$v = \sqrt {2gh}$
Considering $\sqrt 2 = k$ where $k$ is the constant,
$v = k\sqrt {gh}$
Hence proved.