# Motion down an inclined plane

Let P be a particle on a smooth plane inclined at an angle alpha to the horizon. If ABC represent a vertical section of the inclined plane through P, AB is the line of greatest slope and the angle ABC is alpha. The acceleration of P due to gravity is g and it is vertically downwards towards PQ. If PN be normal to the plane then angle QPN is alpha. The resolved parts of g are $g cos\alpha$ along PN and $g sin \alpha$ along PB.

Since the plane prevents any motion perpendicular to it, the particle moves down the plane along AB with acceleration $g sin \alpha$

If h be the height AC and l the length AB of the inclined plane $sin \alpha = \dfrac{h}{l}$

Let a particle slide down AB from rest and v be its velocity at B.

Then $v^2 = 2gsin\alpha l = 2gh$

Thus the velocity acquired is independent of alpha and is the same as the velocity acquired in falling through a height h.

If t be the time taken to slide down AB, $l = \dfrac{1}{2}g sin\alpha t^2$

$t = \sqrt{\dfrac{2l}{g sin\alpha}}$

If the particle is projected upwards along BA, its acceleration is $-g sin\alpha$