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A particle is launched from a point with an initial speed u at an angle θ above the horizontal. At a specific point during its motion, its velocity becomes perpendicular to its original direction. What is the time taken to reach this point?

1. The speed of the particle at this point is u sin θ
2. The speed of the particle at this point is u cos θ
3. The time taken to reach this point is (u/g) sec θ
4. The time taken to reach this point is (u/g) cosec θ

Correct answer: The time taken to reach this point is (u/g) cosec θ

Solution

For velocity to become perpendicular to the launch direction, the component of velocity along the original direction must vanish: u - g t sin(theta) = 0, giving t = u/(g sin theta) = (u/g) cosec theta. The stored 'sec theta' is incorrect; the correct option is (u/g) cosec theta.

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