A bird is flying up at angle ${sin}^{- 1} (3 / 5)$ with the horizontal. A fish in a pond looks at that bird. When it is vertically above the fish. The angle at which the bird appears to fly (to the fish) is $[n_{w a t e r} = 4 / 3]$ .

{sin}^{- 1} (3 / 5)

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{sin}^{- 1} (4 / 5)

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45^{\circ}

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{sin}^{- 1} (9 / 16)

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Solution

The correct option is C $45^{\circ}$
Let $y - a x i s$ be vertically upward and x-axis be horizontal.
$V_{y} (a p p .) = \frac{V_{y} (r e a l)}{(\frac{1}{μ})}$
$V_{y} (a p p .) = V_{x} (r e a l .)$
$tan ϕ = \frac{V_{y} (a p p)}{V_{x} (a p p)} = \frac{4}{3} tan θ = \frac{4}{3} \times \frac{3}{4} = 1$ .

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A bird is flying up at angle sin−1(3/5) with the horizontal. A fish in a pond looks at that bird. When it is vertically above the fish. The angle at which the bird appears to fly (to the fish) is [nwater=4/3].

The correct option is C 45∘Let y−axis be vertically upward and x-axis be horizontal.Vy(app.)=Vy(real)(1μ)Vy(app.)=Vx(real.)tanϕ=Vy(app)Vx(app)=43tanθ=43×34=1.

A bird is flying up at angle ${sin}^{- 1} (3 / 5)$ with the horizontal. A fish in a pond looks at that bird. When it is vertically above the fish. The angle at which the bird appears to fly (to the fish) is $[n_{w a t e r} = 4 / 3]$ .