A projectile is given an initial velocity of (^i+2^j)m/s, where ^i is along the ground and ^j is along the vertical. If g=10m/s2the equation of its trajectory is:
A
y=x−5x2
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B
y=2x−5x2
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C
4y=2x−5x2
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D
4y=2x−25x2
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Solution
The correct option is By=2x−5x2 u=√22+12=√5 θ=tan−12 Equation:y=xtanθ−gx22u2cos2θ y=2x−10x22×5×15 y=2x−5x2