A body is thrown with velocity 20m/s at an angle of 60o with the horizontal. The time gap (in s) between the two positions of body where velocity of body makes an angle of 30o with horizontal is
A
1√3
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B
2√3
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C
√3
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D
3√2
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Solution
The correct option is B2√3
Horizontal component of velocity of particle remains constant during projectile motion. Hence we can equate the horizontal component of velocity of body at point O and A ucos60o=vcos30o v=20√3m/s We can assume the motion of body between points A and B as projectile in horizontal plane Hence, from the formula of time of flight T=2vsin30og=2×20√3×120 T=2√3s