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Question

# A projectile of mass m is launched with an initial velocity →v1 making an angle θ with the horizontal as shown in figure. The projectile moves in the gravitational field to the Earth. The angular momentum of the projectile about the origin when it is at the highest point of its trajectory is:

A
mv3isin2θcosθ2g^k
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B
mv3isin2θcosθ2g^k
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C
mv3isin2θcosθg^k
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D
2mv3isin2θcosθg^k
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Solution

## The correct option is B −mv3isin2θcosθ2g^kAngular momentum =m→r×→vAt the highest point x=12R=u2sin2θ2gy=u2sin2θ2gAngular momentum= L=[u2sin2θ2g^i+u2sin2θ2g^j]×mucosθ^igives, L=mu3sin2θcosθ2g(−^k)

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