    Question

# A particle of mass 5 kg is projected on horizontal ground with an initial velocity of 10 m/s making an angle 30∘ with the horizontal. The magnitude of angular momentum of the particle about the point of projection when the particle is at the highest point is (in kg m2/s) [Take g=10 m/s2]

A
250
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B
1253
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C
zero
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D
12534
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Solution

## The correct option is D 125√34 Given, m=5 kg Initial velocity of particle, u=10 m/s As we know, at maximum height, vertical component of velocity vy=0 and Horizontal component of velocity =vx=ucosθ Hence, angular momentum at highest point →L=m(→r×→v) =mvr⊥(^j×^i)=mucosθ r⊥(−^k) where r⊥=Hmax Hmax=u2sin2θ2g=102×(12)22×10=54 m ∴L=5×(10×cos30∘)×54 →L=125√34(−^k) kgm2s  Suggest Corrections  1      Similar questions  Explore more