A body is thrown with velocity 20m/s at an angle of 60∘ with the horizontal. The time gap between the two positions of body where velocity of body makes an angle of 30∘ with horizontal is
(Take g=10m/s2)
A
1.15s
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B
0.95s
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C
1s
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D
1.5s
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Solution
The correct option is A1.15s
Let A and B be the two points on the trajectory where velocity makes an angle of 30∘ with the horizontal.
Let v1 and v2 be the velocities at point A and B respectively. u=20m/s (given)
Since the horizontal component of velocity always remains constant ⇒v1cos30∘=v2cos30∘=ucos60∘ ⇒v1=v2=u√3 ......(i)
Using first equation of motion for vertical component of velocity from point A to B, we get −v2sin30∘=v1sin30∘−gtAB ......(ii)
Using (i) and (ii) we get gtAB=2×u√3×12=20√3⇒tAB=2√3=1.15s ∴ Time gap between positions A and B=1.15s