The position of a 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 ms-2)

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tan−1⁡23

tan−1⁡32

tan−1⁡74

tan−1⁡45

answer is C.

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Detailed Solution

r¯=40i^+50j^m at t=2secu cos⁡ θt=40 u sin⁡ θt−1/2gt2=50⇒u cos⁡ θ=20 u sin⁡ θ=35⇒u sin ⁡θu cos ⁡θ=3520 tan⁡=7/4⇒θ=tan−1⁡(7/4)

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