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

a
y=2x−5x2
b
y=x−5x2
c
4y=2x−5x2
d
y=2x−25x2

detailed solution

Correct option is A

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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