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loris [4]
3 years ago
9

A child’s toy sits on the bottom of a swimming pool in which the water depth is d=1.8 m. To a child standing at the pool edge, t

he toy appears to be a horizontal distance of 4.2 m away from the sidewall. Assume the child’s eyes are 3.5m above the bottom of the pool. What is the actual horizontal distance from the sidewall to the toy?
Physics
1 answer:
navik [9.2K]3 years ago
8 0

Answer:

Horizontal distance is 1.28 + 2.04 = 3.32 m

Explanation:

Given data:

from  below figure

Applying Pythagoras theorem

from Snell's law

\frac{sin i}{sin r} = \frac{n_2}{n_1} = \frac{\frac{EF}{AB}}{\frac{DE}{BD}}

1.33 =\frac{ \frac{2.04}{\sqrt{(3.5 -1.8)^2 + 2.04^2}}}{\frac{x}{\sqrt{1.8^2 + x^2}}}

solving for x we have

x^2 = 1.626

x = 1.28

Horizontal distance is 1.28 + 2.04 = 3.32 m

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