Question

A body of mass $2\text{kg}$ is projected with a speed $4\text{m/s}$ at an angle ${45}^{\circ }$ to the horizontal. What will be its angular momentum about the point of projection when the body is at the highest point in its trajectory?

• A. $\frac{1.6}{\sqrt{2}}{\text{kg m}}^2/\text{s}$
• B. $\frac{32.6}{\sqrt{2}}{\text{kg m}}^2/\text{s}$
• C. $\frac{16}{\sqrt{2}}{\text{kg m}}^2/\text{s}$
• D. $\frac{3.26}{\sqrt{2}}{\text{kg m}}^2/\text{s}$

Answer 1

The correct option is D $\frac{3.26}{\sqrt{2}}{\text{kg m}}^2/\text{s}$


For given projectile motion,
Range $\left(R\right)=\frac{u^2\sin 2\theta }{g}=\frac{4^2\sin (2\times {45}^{\circ })}{9.8}=\frac{16}{9.8}\text{m}$
Maximum height $\left(H_{max}\right)=\frac{u^2{\sin }^2\theta }{2g}=\frac{4}{9.8}\text{m}$
So at maximum height,
$\overrightarrow{r}=x\hat{i}+y\hat{j}=\frac{R}{2}\hat{i}+H_{max}\hat{j}=(\frac{8}{9.8}\hat{i}+\frac{4}{9.8}\hat{j})\text{m}$
and we know that at highest point velocity will be in $x$ direction only,
$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}=ucos\theta \hat{i}+0\hat{j}=4cos{45}^{\circ }\hat{i}+0\hat{j}=(\frac{4}{\sqrt{2}}\hat{i}+0\hat{j})\text{m/s}$
Now we know,
Angular momentum about origin $\left(\overrightarrow{L}\right)=\overrightarrow{r}\times \overrightarrow{p}$
$=\overrightarrow{r}\times \left(m\overrightarrow{u}\right)=m(\overrightarrow{r}\times \overrightarrow{u})=2[(\frac{8}{9.8}\hat{i}+\frac{4}{9.8}\hat{j})\times (\frac{4}{\sqrt{2}}\hat{i}+0\hat{j})]=2\times \frac{4\times 4}{9.8\times \sqrt{2}}(\hat{j}\times \hat{i})$
$=\frac{3.26}{\sqrt{2}}(-\hat{k})$ as, $(\hat{j}\times \hat{i}=-\hat{k})$
$|L|=\frac{3.26}{\sqrt{2}}{\text{kg m}}^2/\text{s}$
Hence, option (d) is the correct answer.