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Q.

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

(23−1)sec⁡

b

6sec⁡

c

(5−1)sec⁡

d

(6−2)sec⁡

answer is D.

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Detailed Solution

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).