Question

A stone of mass $m$ is projected with a velocity $u$ at an angle of ${45}^o$ to the horizontal. Its angular momentum about the point of projection when it is at its highest point is :

• A. $\frac{m{u}^3}{4\sqrt{2}g}$
• B. $\frac{m{u}^2}{4g}$
• C. $\frac{mu}{\sqrt{2}}$
• D. $\frac{2m{u}^3}{g}$

Answer 1

The correct option is A $\frac{m{u}^3}{4\sqrt{2}g}$

At the highest point , velocity will only be in x direction

i.e. $\overrightarrow{u}=u\cos \theta \hat{i}+0\hat{j}$

while $\overrightarrow{r}=r_x\hat{i}+r_y\hat{j}$

$r_y=\frac{u^2{\sin }^2{45}^0}{2g},r_x=\frac{Range}{2}=\frac{u^2\sin {90}^0}{2g}$

So, Angular momentum about point of projection:

$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$

or, $\overrightarrow{L}=m(r_x\hat{i}+r_y\hat{j})\times \left(u\cos \theta \hat{i}\right)$

or, $\overrightarrow{L}=m{r}_{y}u\cos \theta (-\hat{k})=m\left(u\cos {45}^0\right)\left(\frac{u^2}{4g}\right)$

$=\frac{m{u}^3}{4\sqrt{2}g}.$