A particle is thrown with initial velocity 8 m/s making angle 120∘ with constant acceleration vector 2m/s2 as shown. The time after speed again becomes 8m/s is
A
1s
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B
2s
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C
4s
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D
8s
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Solution
The correct option is C 4s vx=ux+axt=8−(2sin30∘)t=8−t