Physics, 05.08.2019 18:30 montgomerykarloxc24x

Aswimming pool is 1.4 m deep and 12 m long. is it possible for you to dive to the very bottom of the pool so people standing on the deck at the end of the pool do not see you?
part a assume you dive to the very bottom of the pool so people standing on the deck at the end of the pool can see you. find the minimum height of the person's eyes at the far side of the pool, so you would be observable by this person. take an angle 48.5∘ for a ray that is incident at the water-air interface (nwater = 1.33). express your answer to two significant figures and include the appropriate units.
part b therefore, is it possible for you to dive to the very bottom of the pool so people standing on the deck at the end of the pool do not see you?
therefore, is it possible for you to dive to the very bottom of the pool so people standing on the deck at the end of the pool do not see you?

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we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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