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Question

A particle is projected from the bottom of an inclined plane of inclination 30 (with the horizontal) with speed 40 m/s at an angle 60 with the horizontal. The speed of the particle when its velocity vector is parallel to inclined plane is (g=10 m/s2)

A
403 m/s
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B
20 m/s
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C
203 m/s
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D
10 m/s
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Solution

The correct option is A 403 m/s
Here, we choose our axes along and perpendicular to the inclined plane.



So, ux=ucos30o and uy=usin30o
Also the acceleration components will be ax=gsin30o and ay=gcos30o

When velocity vector is along inclined plane we get vy=0
By 1st law of motion, vy=uy+ayt

i.e vy=40sin30gcos30t=0
i.e t=40sin3010cos30=43 s

Hence the speed of the particle will be along xaxis only
vx=ux+axt

vx=40cos30gsin30t
=403210×12×43=203203=403 m/s

Hence option A is the correct answer.

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