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.

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**Step 1.** Assume $v \propto g^a h^b$, i.e. $v = k\,g^a h^b$ where $k$ is a dimensionless constant.

**Step 2.** Dimensions: $[v]=[L^1T^{-1}]$; $[g]=[L^1T^{-2}]$; $[h]=[L^1]$. So the right side has dimensions $[L^1T^{-2}]^a[L^1]^b = [L^{a+b}T^{-2a}]$.

**Step 3.** Equating powers of L and T: $a+b=1$ and $-2a=-1 \Rightarrow a=1/2$; substituting, $b = 1 - 1/2 = 1/2$. …

Q.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.

Concept understanding — Dimensional Analysis

Q.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=kghv = k\sqrt{gh}v=kgh​ where k is a constant.

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Q.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.

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