Question

An object is freely falling under the gravitational force. Its velocity after travelling 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 entire physics quantities other than base quantities can be deduced from the base quantities by-product of different power of base quantities. In the entire physics dimension of the right-hand side and left-hand side quantities are always the same, so if we know the dimension of one of them we can find other quantities in a given equation.

Formula Used: \[[X]={{[M]}^{a}}{{[L]}^{b}}{{[T]}^{c}}\]

Complete step by step answer:
The dimensional conversation is very important because if the formula is not known then the formula deduces by this dimensional conversion method because if only base quantities are known then others all can derive by base quantity.
Here velocity \[v\]; distance travelled \[h\]; gravitational acceleration \[g\] are given and the dimensions of given quantities are given below.
\[[v]={{[M]}^{0}}{{[L]}^{1}}{{[T]}^{-1}}\]
\[[h]={{[M]}^{0}}{{[L]}^{1}}{{[T]}^{0}}\]
\[[g]={{[M]}^{0}}{{[L]}^{1}}{{[T]}^{-2}}\]
In the question also given that \[v\] depends upon gravitational acceleration \[g\] and distance \[h\]. In form of dimensional terms relation can be given by,
\[ [v]\text{ }\alpha \text{ }{{[g]}^{x}}{{[h]}^{y}}\] ……………….…. (1)
\[\Rightarrow {{[M]}^{0}}{{[L]}^{1}}{{[T]}^{-1}}={{({{[M]}^{0}}{{[L]}^{1}}{{[T]}^{-2}})}^{x}}\times {{({{[M]}^{0}}{{[L]}^{1}}{{[T]}^{0}})}^{y}}\]
\[\Rightarrow {{[M]}^{0}}{{[L]}^{1}}{{[T]}^{-1}}\text{ }\alpha \text{ }{{[M]}^{0}}{{[L]}^{x+y}}{{[T]}^{-2x}}\]
Equating the power of the same terms equal to both side,
\[x+y=1\] and \[-2x=-1\]
\[\Rightarrow x=\dfrac{1}{2}\]
and \[y=1-x\]
\[\Rightarrow y=1-\dfrac{1}{2}\]
\[\Rightarrow y=\dfrac{1}{2}\]
Now putting the value of \[x\] and \[y\] in equation (1),
\[\Rightarrow [v]\text{ }\alpha \text{ }{{[g]}^{\dfrac{1}{2}}}{{[h]}^{\dfrac{1}{2}}}\]
\[\Rightarrow [v]\text{ }\alpha \text{ }\sqrt{gh}\]
\[\Rightarrow [v]=k\sqrt{gh}\]
Where \[k\] is a constant.

Additional Information:
 The base quantities are,

Quantities	Unit	Symbol
Length	meter	m
Mass	Kilogram	Kg
Time	second	s
Electric current	ampere	A
Thermodynamic temperature	kelvin	k
Amount of substance	Mole	mol
Luminous intensity	candela	cd

All other quantities made from the multiplication of some of the above quantities.

Note:
(1) Note that this type of conservation is not applicable to the calculation of magnitude.
(2) In the dimension conservation method speed, average speed, change in speed, all are dimensionally the same.
(3) If an equation has several terms then all terms are dimensionally the same.