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A person is swimming 1.1 m beneath the surface of the water in a swimming pool. A child standing on the diving board drops a ball into the pool directly above the swimmer. The swimmer sees the ball dropped from a height of 4.2 m above the water. From what height was the ball actually dropped?

Respuesta :

okpalawalter8

Answer:

The actual height is  [tex]A =3.158 \ m[/tex]

Explanation:

From the question we are told that

   The depth of the person is  [tex]d = 1.1 \ m[/tex]

    The apparent height  is  [tex]D = 4.2 \ m[/tex]

Generally

     The refractive index of water is  [tex]n_w = 1.33[/tex]

      The refractive index of the air is  [tex]n_a = 1[/tex]

The apparent depth is mathematically represented as

      [tex]D = A [\frac{n_w}{n_a} ][/tex]  

substituting values

     [tex]4.2 = A [\frac{1.33}{1} ][/tex]  

=>   [tex]A = \frac{4.2 }{1.33}[/tex]

      [tex]A =3.158 \ m[/tex]

                 

Cricetus

The ball was dropped at the height of "3.158 m". To understand the calculation, check below.

Refractive Index

According to the question,

Water's refractive index, [tex]n_w[/tex] = 1.33

Air's refractive index, [tex]n_a[/tex] = 1

Apparent height, D = 4.2 m

Person's depth, d = 1.1 m

We know the relation,

→ D = A[[tex]\frac{n_w}{n_a}[/tex]]

By substituting the values, we get

4.2 = A[[tex]\frac{1.33}{1}[/tex]]

By applying cross-multiplication,

  A = [tex]\frac{4.2}{1.33}[/tex]

      = 3.158 m

Thus the approach above is correct.

Find out more information about refractive index here:

https://brainly.com/question/10729741