JEE Main · 2005 · 08 May 2005 · Offline · Chapter_pending

A particle is projected from a point O with velocity

u

at an angle of

6 0^{\circ}

with the horizontal. When it is moving in a direction at right angles to its direction at O, its velocity then is given by

Answer: (D)

Solution

In projectile motion the horizontal component of velocity stays constant, equal to

u cos 6 0^{\circ} = \frac{u}{2}

. When the velocity becomes perpendicular to the initial direction, it points at

3 0^{\circ}

below the horizontal, since the initial direction was

6 0^{\circ}

above it. If the speed then is

v

, matching horizontal components gives

v cos 3 0^{\circ} = u cos 6 0^{\circ}

, so

v \cdot \frac{3}{2} = \frac{u}{2}

, and

v = \frac{u}{3}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option