The velocity of a freely falling body changes as $g^{p}h^{q}$ where $g$ is acceleration due to gravity and $h$ is the height. The values of $p$ and $q$ are

Question

Dr. Abubacker · Accepted Answer

$v \propto g^{p}h^{q}\$(given) By submitting the dimension of each quantity and comparing the powers on both sides we get $\left\lbrack LT^{- 1} ightbrack = \left\lbrack LT^{- 2} ightbrack^{p}\lbrack Lbrack^{q}$ $\Rightarrow p + q = 1,\ - 2p = - 1,\ 	herefore p = \frac{1}{2},\ q = \frac{1}{2}$

Q. The velocity of a freely falling body changes as $g^{p}h^{q}$ where $g$ is acceleration due to gravity and $h$ is the height. The values of $p$ and $q$ are

Explanation

Reviewed by Dr. Abubacker

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