A projectile is projected at $30^{\circ}$ from horizontal with initial velocity $40 \mathrm{~ms}^{-1}$. The velocity of the projectile at $\mathrm{t}=2 \mathrm{~s}$ from the start will be : (Given $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )

To find the velocity of the projectile at t = 2 s, we need to find the horizontal and vertical components of the velocity at that time. <br/><br/> The initial horizontal component of the velocity is constant and is given by: <br/><br/> $$ v_{0x} = v_0 \cos\theta = 40\,\mathrm{ms}^{-1} \cos(30^{\circ}) = 40\,\mathrm{ms}^{-1} \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}\,\mathrm{ms}^{-1} $$ <br/><br/> The initial vertical component of the velocity is: <br/><br/> $$ v_{0y} = v_0 \sin\theta = 40\,\mathrm{ms}^{-1} \sin(30^{\circ}) = 40\,\mathrm{ms}^{-1} \cdot \frac{1}{2} = 20\,\mathrm{ms}^{-1} $$ <br/><br/> To find the vertical component of the velocity at t = 2 s, we use the equation: <br/><br/> $$ v_y = v_{0y} - gt = 20\,\mathrm{ms}^{-1} - (10\,\mathrm{ms}^{-2})(2\,\mathrm{s}) = 20\,\mathrm{ms}^{-1} - 20\,\mathrm{ms}^{-1} = 0\,\mathrm{ms}^{-1} $$ <br/><br/> At t = 2 s, the horizontal component of the velocity is still $20\sqrt{3}\,\mathrm{ms}^{-1}$, and the vertical component is 0. The overall velocity at t = 2 s is: <br/><br/> $$ \vec{v} = 20\sqrt{3}\,\mathrm{ms}^{-1} \hat{i} + 0\,\mathrm{ms}^{-1} \hat{j} = 20\sqrt{3}\,\mathrm{ms}^{-1} $$ <br/><br/> So the correct answer is: $20\sqrt{3}\,\mathrm{ms}^{-1}$