Kinematics

JEE Main level, single correct, medium

219 of 308

Question 219 of 308

A particle of mass mm projected with a velocity uu making an angle of 30∘30^{\circ} with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height hh is :

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Answer: (A) 316mu3g\dfrac{\sqrt{3}}{16}\dfrac{m{u}^{3}}{g}

Angular momentum at maximum height will be, L=(mucos⁡θ)hL=\left(mu\cos \theta \right)h, where h=h=maximum height
Therefore, L=mucos⁡θ×u2sin⁡2θ2gL=mu\cos \theta \times \dfrac{{u}^{2}{\sin }^{2}\theta }{2g}
=mu32g×32×(12)2=3mu316g=\dfrac{m{u}^{3}}{2g}\times \dfrac{\sqrt{3}}{2}\times {\left(\dfrac{1}{2}\right)}^{2}=\dfrac{\sqrt{3}m{u}^{3}}{16g}

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