Question

Check the correctness of given physical relation MGH equal one by two m vsquare

Answer 1

$\sf\bf\underline{\underline{\bigstar\huge ANSWER\bigstar}}$

Equation ,

$\sf \longrightarrow MGH=\dfrac{1}{2}mv^2$

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L.H.S = MGH

Here ,

M = Mass

G = Acceleration due to gravity

H = Height

$\sf MGH = [M^1L^0T^0][M^0L^1T^-2][M^0L^1T^0]$

        $\sf = [M^1L^2T^-2]$

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R.H.S = $\sf\dfrac{1}{2}mv^2$

Here ,

$\sf \dfrac{1}{2}$ = constant

M = Mass

V = Velocity

$\sf mv^2= [M^1L^0T^0][M^0L^1T^-2]^2$

      $= \sf [M^1L^0T^0][M^0L^2T^-2]$

      $= \sf [M^1L^2T^-2]$

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$\bf\therefore \ L.H.S=R.H.S$

HOPE IT HELPS U ...... :)