A diver D is still under water μ$$=43$$ at a depth d $$=$$ $$10$$ m. A bird is diving along line AB at a constant velocity in air. When the bird is exactly above the diver he sees it ata height of $$50$$ m from himself and velocity of the bird appears to be inclined at $$45$$° to the horizontal. At what distance (in$$ m $$) from the diver the bird actually hits the water surface.

Question

A diver D is still under water μ$$=43$$ at a depth d $$=$$ $$10$$ m. A bird is diving along line AB at a constant velocity in air. When the bird is exactly above the diver he sees it ata height of $$50$$ m from himself and velocity of the bird appears to be inclined at $$45$$° to the horizontal. At what distance (in$$ m $$) from the diver the bird actually hits the water surface.

Infinity Learn · Accepted Answer

$$happ=$$μh    $$happ=50$$−$$d=40m$$ and $$m=$$μ$$=4/3$$∴    $$h=30m$$ Now,     happ $$=$$μh    Vy app $$=$$μVy    Vx app $$=Vx$$ Given     Vyapp Vxapp $$=$$$$tan$$⁡$$45$$∘$$=1$$∴     μ$$VyVx=1$$⇒$$tan$$⁡α$$=1$$μ$$=34$$α$$=37$$∘α is the true angle that the line of motion of the bird makes with horizontal. ∴ $$OC=htan$$⁡α$$=30$$⋅$$34=22.5m$$∴ $$DC=(22.5)2+102=24.62m$$

Refraction of light

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Refraction of light

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happ=μh happ=50−d=40m and m=μ=4/3∴ h=30m Now, happ =μh Vy app =μVy Vx app =Vx Given Vyapp Vxapp =tan⁡45∘=1∴ μVyVx=1⇒tan⁡α=1μ=34α=37∘α is the true angle that the line of motion of the bird makes with horizontal. ∴ OC=htan⁡α=30⋅34=22.5m∴ DC=(22.5)2+102=24.62m

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