Question

7. The distance travelled by a freely falling body is given by the equation, y = v0 t – 1 2 gt2 . Using principles of dimensions, check the correctness of the equation.​

Answer 1

Answer:

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Answer 2

Given : The distance travelled by a freely falling body is given by the equation,

y = $\[{v_0}t - \frac{1}{2}g{t^2}\]$

To Check : Using principles of dimensions, check the correctness of the equation.​

Explanation:

Writing the dimensions of either side of the given equation.

y = $\[{v_0}t - \frac{1}{2}g{t^2}\]$

LHS= y =displacement = [$\[{M^0}{L^1}{T^0}\]$]

RHS = t = velocity×time = [$\[{M^0}{L^0}{T^{ - 1}}\]$][T] = [$\[{M^0}{L^0}{T^0}\]$ ]

and

 $\[\frac{1}{2}g{t^2}\]$ = $\[acceleration \times {(time)^2}\]$ =[$\[{M^0}{L^0}{T^{ - 2}}\]$ ]$\[{[T]^2}\]$ = [$\[{M^0}{L^0}{T^0}\]$ ]

As LHS=RHS

Hence the formula is dimensionally correct.