Question

An object is falling freely under gravitational force. Its velocity after travelling a distance h is v. If v depends upon gravitational acceleration g and distance, prove with the dimensional analysis that v = k`sqrt"gh"` where k is a constant.

Answer 1

Given = v = k`sqrt"gh"`

v ∝ gh

v = kg^ah^b   ...(i)

Quantity	Dimension
Velocity (v)	[L1 M0 T-1]
Height (h)	[L1 M0 T0]
Gravitational acceleration (g)	[L1 M0 T-2]

k being constant is assumed to be dimensionless.

v = k g^a h^b 

[L¹ M⁰ T^-1] = k `["L"^1 "M"^0 "T"^-2]^"a" ["L"^1 "M"^0 "T"^0]^"b"`

[L¹ M⁰ T^-1] = k `["L"^"a" "M"^0 "T"^ -"2a"] ["L"^"b" "M"^0 "T"^0]`

[L¹ M⁰ T^-1] = k `["L"^("a + b") "M"^0 "T"^(- 2"a")]`

∴ a + b = 1  ...(ii)

∴ - 2a = - 1

⇒ a = `bb(1/2)`

Put a = `1/2` in equation (ii)

∴ a + b = 1

⇒ `1/2` + b = 1

⇒ b = `1 - 1/2`

⇒ b = `2/2 -1/2`

⇒ b = `bb(1/2)`

∴  v = k g^a h^b 

v = k `"g"^(1/2) "h"^(1/2)`

v = k`sqrt"gh"`