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Projectile Basics & Trajectory question from JEE Main 2013

Q7JEE Main 2013Easy

Projectile Basics & Trajectory

A projectile is given an initial velocity of (i^+2j^)\left(\hat{i} + 2\hat{j}\right) m/s, where i^\hat{i} is along the ground and j^\hat{j} is along the vertical. If g = 10 m/s2, the equation of its trajectory is:

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Answer: (b)

  1. ux=1u_x = 1, uy=2u_y = 2: x=tx = t, y=2t−5t2y = 2t - 5t^2
  2. Put t=xt = x: y=2x−5x2y = 2x - 5x^2
  3. So the answer is (b).
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