Question

A particle has mass $M=1kg$ . It is projected with a velocity $u=5m/s$ at an angle $\theta ={60}^{\circ }$ with the horizontal. The particle has only a horizontal component of velocity at the highest point (Refer image). When the particle reaches the highest point of its trajectory calculate the angular momentum of the particle about the point of projection.
(Take $g=10m/s^2$)

• A. $\frac{75}{32}N-m-s$
• B. $\frac{32}{75}N-m-s$
• C. $\frac{75}{36}N-m-s$
• D. $\frac{36}{75}N-m-s$

Answer 1

The correct option is A $\frac{75}{32}N-m-s$

At the highest point the horizontal velocity is $u_0=u\cos {60}^{\circ }=5\times \frac{1}{2}=2.5m/s$

The perpendicular distance from point O

$H=\frac{u^2{\sin }^2\theta }{2g}$

$=\frac{5^2.{\sin }^{2}60}{2\times 10}$

$=\frac{25}{20}\times {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{15}{16}m$

Required Angular Momentum $=m{u}_0\times H$

$\Rightarrow L=1\times 2.5\times \frac{15}{16}$
$\Rightarrow L=\frac{75}{32}N-m-s$

Correct Option - (A)