Question

A ball of mass $m=2\text{kg}$ is projected at an angle $\theta ={45}^{\circ }$ with initial velocity of $u=10\text{m/s}$ as shown in figure. Find the torque acting on the particle due to gravity about origin at it's maximum height.
(Take $g=10m/s^2$)

• A. $100\hat{k}$
• B. $10\hat{k}$
• C. $-100\hat{k}$
• D. $-10\hat{k}$

Answer 1

The correct option is C $-100\hat{k}$


We know,
Range of projectile$R=\frac{u^2\sin \left(2\theta \right)}{g}=\frac{{10}^2\times \sin (2\times {45}^{\circ })}{10}=10\text{m}$
Maximum height$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{{10}^2{\sin }^2{45}^{\circ }}{2\times 10}=2.5\text{m}$


When ball reaches to maximum height $H$
$\overrightarrow{r}=\frac{R}{2}\hat{i}+H\hat{j}=(5\hat{i}+2.5\hat{j})\text{m}$
Force due to gravity$\overrightarrow{F}=-mg=-2\times 10\text{N}=-20\hat{j}\text{N}$

Torque due to gravity about origin$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=(5\hat{i}+2.5\hat{j})\times -20\hat{j}=-100\hat{k}\text{Nm}$