Question

Check the correctness of the eqation
H=(v^2sin^2 thetta ) / 2 g. where H= maximum height v= initial velocity theetta= angle of projection. g = acceleration due to gravity?
plz help me​
question belogs to +1 physics

Answer 1

To check the correctness of the equation, we have to use dimensional analysis. If the dimensions of L.H.S. and R.H.S. are same, that means the equation in dimensionally correct.

L.H.S. :-

$\underline{\underline{\green{\sf{H = \bigg\{ L\bigg\} }}}}$

R.H.S :-

$\sf \dfrac{v^2 sin^2\theta}{2g}$

As $\sf sin\theta$ is dimensionless, $\sf sin^2\theta$ will also be dimensionless.

$\sf \Bigg\{\dfrac{v^2 sin^2\theta}{2g}\Bigg\} = \Bigg\{ \dfrac{ \Big( \scriptsize\dfrac{L}{T}\Big)^2 }{ \Big( \scriptsize\dfrac{L}{T^2 }\Big) } \Bigg\}$

$\longrightarrow \sf \Bigg\{\dfrac{v^2 sin^2\theta}{2g}\Bigg\} = \Bigg\{ \dfrac{ \Big( \scriptsize\dfrac{L^2}{T^2}\Big)}{ \Big( \scriptsize\dfrac{L}{T^2}\Big) }\Bigg\}$

$\longrightarrow \sf\Bigg\{ \dfrac{v^2 sin^2\theta}{2g}\Bigg\} =\Bigg\{ \dfrac{ \Big( \scriptsize\dfrac{L^2}{{\cancel{T^2}}}\Big)}{ \Big( \scriptsize\dfrac{L}{{\cancel{T^2}}}\Big)} \Bigg\}$

$\longrightarrow \sf \Bigg\{\dfrac{v^2 sin^2 \theta}{2g}\Bigg\} = \Bigg\{ \dfrac{L^2}{L} \Bigg\}$

$\longrightarrow \sf\Bigg\{ \dfrac{v^2 sin^2 \theta}{2g}\Bigg\} = \Bigg\{\dfrac{ {L}^{ {\cancel{2}}} }{ {\cancel{L}} } \Bigg\}$

$\longrightarrow \underline{\underline{\green{\sf{\Bigg\{ \dfrac{v^2 sin^2 \theta}{2g}\Bigg\} =\Bigg\{ L \Bigg\} }}}}$

As the dimensions of L.H.S. and R.H.S. are same, the equation in dimensionally correct.