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/s2, the equation of its trajectory is:
A
y=2x−5x2
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B
4y=2x−5x2
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C
4y=2x−25x2
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D
y=x−5x2
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Solution
The correct option is Cy=2x−5x2 →u=^i+2^j; ∴ucosθ=1;usinθ=2 x=ucosθt=t y=usinθt−12gt2=2t−5t2
Substituting t=x Equation of trajectory is y=2x−5x2