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?

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-06-10T02:37:12+02:00

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 Cricetus · Answer 2 · 2022-04-02T16:30:16+02:00

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

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