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A particle is projected from ground with a velocity of $20 \sqrt{2} m / s$ making an angle of $60^{o}$ with horizontal. Then the time after which its velocity vector makes an angle of $45^{o}$ with horizontal is (Take $g = 10 m / s^{2}$ ).

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(23−1)sec⁡

6sec⁡

(5−1)sec⁡

(6−2)sec⁡

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detailed solution

Correct option is D

Horizontal component of velocity remains constant.∴ Vcos⁡45o=ucos⁡60o⇒V.12=202.12⇒V=20 m/sNow, Vsin⁡45o=usin⁡60o−gt⇒t=202×32−20×1210sec⁡=(6−2)sec⁡

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A particle is projected from ground with a velocity of 202 m/s making an angle of 60o with horizontal. Then the time after which its velocity vector makes an angle of 45o with horizontal is (Take g=10 m/s2).

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A particle is projected from ground with a velocity of $20 \sqrt{2} m / s$ making an angle of $60^{o}$ with horizontal. Then the time after which its velocity vector makes an angle of $45^{o}$ with horizontal is (Take $g = 10 m / s^{2}$ ).