A projectile is given an initial velocity of (2^i+4^j)m/s, where ^i is along the ground and ^j is along the vertical. If g=10m/s2, then the equation of its trajectory is
A
y=x−5x2
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B
y=2x−54x2
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C
y=4x−5x2
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D
y=4x−25x2
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Solution
The correct option is By=2x−54x2 Initial velocity = 2^i+4^j
Magnitude of initial velocity = u=√22+42=√20m/s
If θ be the angle of projectile, tanθ=uyux=42=2
Equation of trajectory is given by y=xtanθ−gx22u2(1+tan2θ) y=x×2−10x22×20(1+4) ∴y=2x−54x2