A projectile is given an initial velocity of i^+2j^ . The Cartesian equation of its path is: (g=10 m/s2)

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y=2x−5x2

y=x−5x2

4y=2x−5x2

y=2x−25x2

answer is A.

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Detailed Solution

u→=uxi^+uyj^=ucosθi^+usinθj^=i^+2j^∴ tanθ=usinθucosθ=21 , cosθ=15The desired equation is,y=xtanθ−gx22u2cos2θ=x×2−10x22(22+12)2(15)2or y=2x−5x2

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