IIT JEE , AIEEE Forum - Physics , Mechanics

neeraj

Let us consider an inclined plane which makes an angle b with the horizontal. Let a particle projected at angle a with horizontal with a velocity u. This particle strikes an inclined plane at a point A. Our aim is to find out time of flight and range OA

The initial velocity can be resolved into two components

(i) u cos (a - b) along the plane

(ii) u sin (a - b) perpendicular to the plane

Similarly g can be resolve in two components

(i) g sinb parallel to the plane (retardation)

(ii) g cosb perpendicular to the plane

If t be the time taken by the particle to go from O to A then in this time the distance described perpendicular to OA is zero

S = u t + ½ g t²

0 = u sin (a - b) - ½ g cos b. t²

t = [ 2 u sin (a - b)] / [g cos b ] -----.(1)

eq.1 represents time of flight

During time t horizontal velocity u cos a along OB remains const.

Distance traveled OB is given by

OB = ( u cos a) . t

= (u cos a) [ 2 u sin (a - b)] / g cos b

= [ 2 u² sin (a - b) . cos a ] / g cos b ---.. (2)

OA = OB / cos b = = [ 2 u² sin (a - b) . cos b ] / g cos² b

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