Question

A particle is projected at an angle of ${30}^{\circ }$ w.r.t horizontal with speed $20m/s.$ Take the point of projection as the origin of the coordinate axes and find the position vector(in $m$) of the particle after $1s$. (Take $g=10m/s^2$)

• A. $5\sqrt{3}\hat{i}+5\hat{j}$
• B. $10\sqrt{3}\hat{i}+5\hat{j}$
• C. $5\hat{i}+10\sqrt{3}\hat{j}$
• D. $10\sqrt{3}\hat{i}+10\hat{j}$

Answer 1

The correct option is B $10\sqrt{3}\hat{i}+5\hat{j}$

Displacement of the particle $x$-direction is given by
$x=u\cos \theta \times t=20\times \frac{\sqrt{3}}{2}\times 1=10\sqrt{3}m$
Displacement of the particle in $y$-direction is given by,
$y=u\sin \theta t-\frac{1}{2}\times 10\times t^2$
$=20\times \frac{1}{2}\times 1-5\times 1^2=5m$
$\therefore$ Position vector $\left(\overrightarrow{r}\right)=10\sqrt{3}\hat{i}+5\hat{j}$