Question

A particle is projected at an angle of ${30}^{\circ }$ with respect to the horizontal with speed $20m/s$. Find the position vector of the particle after $1s$.

• A. $\sqrt{(10\sqrt{3}{)}^2+5^2}$
• B. $\sqrt{(10\sqrt{3}{)}^2-5^2}$
• C. $\sqrt{(10\sqrt{3}{)}^2+5}$
• D. $\sqrt{(10\sqrt{3}{)}^2-5}$

Answer 1

The correct option is A $\sqrt{(10\sqrt{3}{)}^2+5^2}$

Given : $u=20m/s$ $\theta ={30}^o$ $t=1s$

Horizontal direction :

Initial speed $u_x=u\cos {30}^o=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}m/s$

Acceleration $a_x=0$

Displacement in horizontal direction $S_x=u_{x}t=10\sqrt{3}\times 1=10\sqrt{3}m$

Vertical direction :

Considering upward direction to be negative :

Initial speed $u_y=-u\sin {30}^o=-20\times \frac{1}{2}=-10m/s$

Acceleration $a_y=g=10m/s^2$

Displacement in vertical direction $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2=-10\times 1+\frac{10}{2}(1{)}^2=-5m$

Thus net displacement $S=10\sqrt{3}\hat{i}-5\hat{j}$

Position of body after 1 second $|S|=\sqrt{(10\sqrt{3}{)}^2+5^2}$