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Question 5360
NEET Physics · Hard

Motion in a Plane: Mixed 2D Motion MCQ

Solve this quality-checked NEET multiple-choice question, then review the correct answer and explanation.

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Mixed 2D MotionHardQuestion 5360

Question

A projectile is launched at 45° with speed u. The time at which the velocity vector makes 30° with horizontal is:
  1. A
    u(tan45°−tan30°)/g
  2. B
    u(1−1/√3)/g
    Correct
  3. C
    u/g
  4. D
    u/(2g)

Correct answer

u(1−1/√3)/g

Explanation

At angle 30°: vₓ = ucosθ = u/√2, vy = vₓtan30° = u/(√2√3) = u/√6. But initially vy = u/√2. vy = u/√2 − gt. At 30°: vy/vₓ = tan30° = 1/√3. vy = vₓ/√3 = (u/√2)/√3 = u/√6. So u/√6 = u/√2 − gt → t = (u/√2 − u/√6)/g = u(1/√2 − 1/√6)/g = u(1−1/√3)/g (factoring out 1/√2).

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