The position of projectile launched from the origin at t=0 is given by r→=40 i^+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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a
tan−12/3
b
tan−13/2
c
tan−17/4
d
tan−14/5
answer is C.
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Detailed Solution
The position vector of the projectile is r→=u→t+12a→t240 i^+50j^=u→×2+12−10j^×22⇒u→=uxi^+uyj^=20 i^+35j^tanθ=uyux=3520=74⇒θ=tan−174