Question

A particle of mass $5\text{kg}$ is projected on horizontal ground with an initial velocity of $10\text{m/s}$ making an angle ${30}^{\circ }$ 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 ${\text{kg m}}^2\text{/s}$) [Take $g=10{\text{m/s}}^2$]

• A. $250$
• B. $125\sqrt{3}$
• C. zero
• D. $\frac{125\sqrt{3}}{4}$

Answer 1

The correct option is D $\frac{125\sqrt{3}}{4}$


Given, $m=5\text{kg}$
Initial velocity of particle, $u=10\text{m/s}$
As we know, at maximum height, vertical component of velocity $v_y=0$
and Horizontal component of velocity $=v_x=u\cos \theta$

Hence, angular momentum at highest point
$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$
$=mv{r}_{\perp }(\hat{j}\times \hat{i})=mu\cos \theta r_{\perp }(-\hat{k})$
where $r_{\perp }=H_{max}$
$H_{max}=\frac{u^2{\sin }^2\theta }{2g}=\frac{{10}^2\times {\left(\frac{1}{2}\right)}^2}{2\times 10}=\frac{5}{4}\text{m}$
$\therefore L=5\times (10\times \cos {30}^{\circ })\times \frac{5}{4}$
$\overrightarrow{L}=\frac{125\sqrt{3}}{4}(-\hat{k})\text{kg}\frac{m^2}{s}$