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Q. A particle is projected at point A from an inclined plane with inclination angle $\theta$ as shown in figure. The magnitude of projection velocity is $\vec{u}$ and its direction is perpendicular to the plane. After some time it passes from point B which is in the same horizontal level of A, with velocity $\vec{v}$ . Then the angle between $\vec{u}$ and $\vec{v}$ will be
image

Motion in a Plane

Solution:

$\alpha$ is the required angle.
image
so $\alpha=180^{\circ}-2\theta$