Question

A particle of mass $m=1\text{kg}$ is thrown from a height of $20\text{m}$ from the ground with a horizontal speed $10\text{m/s}$. Find the angular momentum [in ${\text{kg m}}^2\text{/s}$] of the particle about the point of projection when it hits the ground. (Take $g=10{\text{m/s}}^2$)

• A. Zero
• B. $400\left(\hat{k}\right)$
• C. $-200\left(\hat{k}\right)$
• D. $600\left(\hat{k}\right)$

Answer 1

The correct option is C $-200\left(\hat{k}\right)$

Initially, at time $t=0$
Horizontal component of speed $u_x=u=10\text{m/s}$
Vertical component of speed $u_y=0$
At time $t$, when it hits the ground:
(Assume upward motion as positive)
Horizontal speed $v_x=u_x$ (no horizontal acceleration)
From eq. $v^2=u^2+2as$,
$v_y^2=u_y^2+2a_{y}s=0-2gs$
$\Rightarrow v_y^2=2\times (-10)\times (-20)$
or $v_y=20\text{m/s}$
So velocity vector when particle hits the ground,
$\overrightarrow{v}=10\hat{i}-20\hat{j}=v_x\hat{i}+v_y\hat{j}$
Time of flight $t$ can be calculated from,
$v_y=u_y+a_{y}t\Rightarrow 20=0+10\times t$
i.e $t=2\text{s}$
So, distance travelled by the particle:
In horizontal direction, $S_x=u_{x}t=20\text{m}$
In vertical direction $S_y=20\text{m}$
Hence, particle will hit the ground at $A(20,0)$.


Position vector from point of projection,
$\overrightarrow{r}=(20-0)\hat{i}+(0-20)\hat{j}=20\hat{i}-20\hat{j}$
As we know, angular momentum
$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$
$\overrightarrow{L}=1[(20\hat{i}-20\hat{j})\times (10\hat{i}-20\hat{j})]$
$=-400(\hat{i}\times \hat{j})-200(\hat{j}\times \hat{i})$
$=-400\hat{k}-200(-\hat{k})$
$\overrightarrow{L}=-200\hat{k}$