Physics

$$V\propto { g }^{ x }{ h }^{ x }$$, where $$V$$ is velocity, $$g$$ is  acceleration due to gravity and $$h$$ is height. Then $$x$$ and $$y$$ are:


ANSWER

$$\dfrac { 1 }{ 2 } ,\dfrac { 1 }{ 2 } $$


SOLUTION

$$V  \propto g^{x} h^{y}$$

$$LT^{-1}=(LT^{-2})^{x}(L)^{y}$$

$$L^{1}T^{-1}=L^{x+y}T^{-2x}$$

$$x+y=1$$

$$-2x=-1$$
$$ \Rightarrow x=\dfrac{1}{2}, y=\dfrac{1}{2}$$

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