The position of projectile launched from the origin at t = 0 is given by r→=40i^+50j^m at t = 2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 m/s2).

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tan-1(2/3)

tan-1(3/2)

tan-1(7/4)

tan-1(4/5)

answer is C.

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Detailed Solution

The position vector of the projectile is r→=ut→+12at→240i^+50j^=u→×2+12(−10j^)×22⇒u→=uxi^+uyj^=20i^+35j^tan⁡θ=uyux=3520=74⇒θ=tan−1⁡74

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