Question

A projectile is projected from the origin in the $x-y$ plane with a velocity of $60\text{m/s}$ at a ${30}^{\circ }$ with the horizontal. Find the position vector of the projectile after $2\text{s}$. Take $(g=10{\text{m/s}}^2)$

• A. $40\hat{i}+40\hat{j}$
• B. $60\sqrt{3}\hat{i}+40\hat{j}$
• C. $40\hat{i}+60\sqrt{3}\hat{j}$
• D. $60\hat{i}+60\hat{j}$

Answer 1

The correct option is B $60\sqrt{3}\hat{i}+40\hat{j}$


For $x-$ component
$s_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$\Rightarrow s_x=60\cos 30°\left(2\right)+0$ $(\because a_x=0)$
$\Rightarrow s_x=60\times \frac{\sqrt{3}}{2}\times 2=60\sqrt{3}$
For $y-$ component
$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow s_y=u\sin \theta \left(2\right)+\frac{1}{2}(-10)(2{)}^2$
$\Rightarrow s_y=60\left(\frac{1}{2}\right)\left(2\right)+(-20)=40\text{m}$
$\therefore$ Position vector is $60\sqrt{3}\hat{i}+40\hat{j}$