A particle is projected with a certain velocity at an angle α above the horizontal from the foot of an inclined plane of inclination 30°. If the particle strikes the plane normally, thenα is equal to
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a
30°+tan-132
b
45°
c
60o
d
30°+tan-1(23)
answer is A.
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Detailed Solution
tAB= time of flight of projectile =2usinα-30°gcos30° Now component of velocity along the plane becomes zero at point B. 0=ucosα-30°-gsin30°×T ucosα-30°=gsin30°×2usinα-30°gcos30° tanα-30°=cot30°2=32 α=30°+tan-132