Saved Bookmarks
| 1. |
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 \( v =k \sqrt{g h} \) where \( k \) is \( a \) constant. |
|
Answer» Given \(v = k\sqrt{ gh}\) Dimensional formula \(v = [LT^{-1}]\) Dimensional formula \(\sqrt {gh} = [LT^{-1} ]^\frac12 [L] ^\frac 12\) \(= \sqrt{[LT^{-2}] [L]}\) \(= \sqrt{[L^2T^{^{-2}}]}\) \(\sqrt{gh} = \sqrt{[LT^{-1}]^2}\) \(\sqrt{gh} = [LT^{-1}]\) Then \([LT^{-1}] = K [LT^{-1}]\) \(k = \frac{[LT^{-1}]}{[LT^{-1}]}\) \(k = 0\) K is dimensionless. |
|