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Test dimensionally if the v^2=u^2+2ax ma...

Test dimensionally if the `v^2=u^2+2ax` may be correct.

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We have `v^(2) - u^(2) = 2as`. Checking the dimensions on both sides , we get
`LHS = [ LT^(-2)]^(2) - [LT^(-1)]^(2) = [L^(2)T^(-2)] - [L^(2)T^(-2) ] - = [L^(2) T^(-2)]`
`RHS = [L^(1)T^(2)] [L] = [L^(2) T^(2)]`
Comparing `LHS and RHS` , we find `LHS = RHS`.
Hence, the formula is dimensionally correct.
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