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Question

A body is thrown with velocity 20 m/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=10 m/s2)

A
1.15 s
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B
0.95 s
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C
1 s
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D
1.5 s
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Solution

The correct option is A 1.15 s

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=20 m/s (given)
Since the horizontal component of velocity always remains constant
v1cos30=v2cos30=ucos60
v1=v2=u3 ......(i)
Using first equation of motion for vertical component of velocity from point A to B, we get
v2sin30=v1sin30gtAB ......(ii)
Using (i) and (ii) we get
gtAB=2×u3×12=203tAB=23=1.15 s
Time gap between positions A and B =1.15 s

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