A projectile of mass m is launched with an initial velocity vi→ making an angle θ with the horizontal as shown in figure. The projectile moves in the gravitational field of the Earth. The angular momentum of the projectile about the origin when it is at the highest point of its trajectory is
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a
-mvi3 sin2θ cos θ2gk^
b
mvi3sin2θ cos θ2gk^
c
-mvi3sin θ2cos θgk^
d
2mvi3sin θ2 cos θ gk^
answer is A.
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Detailed Solution
At the highest point of the trajectory,x = 12R = vi2sin 2θ2g and y = hmax = (vi sin θ)22gThe angular momentum is thenL1→ = r1→×mv1→ = [vi2sin 2θ2gi^+(visin θ)22gj^]×mvxii^ = -mvi3sin2θcos θ2gk^