Question

Using dimensional analysis, verify the correctness of
1. v = 2gh and
2. v = ut/2 + at^2

Answer 1

In dimensional analysis, an equation is correct if the dimension of its left-hand side (LHS) and right-hand side (RHS) are equal, that is

$$[LHS] = [RHS].$$

The dimension of variables:

final velocity

$$v = [LT^{-1}],$$

initial velocity

$$u=[LT^{-1}],$$

acceleration

$$g=[LT^{-2}],$$

height

$$h=[L],$$

time

$$t=[T],$$

number 2 has no dimension, where L is length and T is time.

Solving the problems.

1. For a given expression 

$$v = 2gh$$

true that

$$LHS = v, [LHS] = [LT^{-1}];$$ $$RHS = 2gh, [RHS] = [LT^{-2}][L] = [L^2 T^{-2}];$$ $$[LT^{-1}]\neq[L^2 T^{-2}],$$

that is

$$[LHS]\neq[RHS].$$

Last inequality means that expression isn`t correct.

2. 

$$v = \frac{ut}{2} + at^2;$$ $$LHS = v, [LHS] = [LT^{-1}];$$ $$RHS = \frac{ut}{2} + at^2,$$ $$[RHS] = [L T^{-1}][T]+[L T^{-2}][T^2] = [L]+[L] = [L];$$ $$[LHS]\neq[RHS],$$

therefore

$$[LT^{-1}]\neq[L],$$

so, the equation is false.