V^2=u^2+2asShow that this equation is dimensionaly correct

1.	V^2=u^2+2asShow that this equation is dimensionaly correct
Answer» To show that v² = u² + 2as is dimensionally correct $\sf{Here}\begin{cases}\sf{u,v \longrightarrow Velocity \longrightarrow LT^{-1}}\\ \sf{a \longrightarrow Acceleration \longrightarrow LT^{-2}}\\\sf{s \longrightarrow Distance \longrightarrow L}\end{cases}$ By DIMENSIONAL Analysis LHS $\sf v^2 = [LT^{-1}]^2 \\ \\ \longmapsto \sf v^2 = L^2 T^{-2}-----------(1)$ RHS $\sf u^2 + 2as = [LT^{-1}]^2 + [LT^{-2}][L] \\ \\ \longmapsto \sf u^2 + 2as = L^2T^{-2} + L^2T^{-2} --------------(2)$ By Principle Of Homogeneity,I conclude EQUATIONS (1) and (2) are dimensionally SIMILAR THUS, v² = u² + 2as is dimensionally correct

Answer»

To show that v² = u² + 2as is dimensionally correct

$\sf{Here}\begin{cases}\sf{u,v \longrightarrow Velocity \longrightarrow LT^{-1}}\\ \sf{a \longrightarrow Acceleration \longrightarrow LT^{-2}}\\\sf{s \longrightarrow Distance \longrightarrow L}\end{cases}$