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Question

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=10 m/s2, then the equation of its trajectory is

A
y=x5x2
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B
y=2x54x2
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C
y=4x5x2
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D
y=4x25x2
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Solution

The correct option is B y=2x54x2
Initial velocity = 2^i+4^j
Magnitude of initial velocity = u=22+42=20 m/s
If θ be the angle of projectile,
tanθ=uyux=42=2
Equation of trajectory is given by
y=xtanθgx22u2(1+tan2θ)
y=x×210x22×20(1+4)
y=2x54x2

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