QJEEM23Ph8342: JEE Main 2023 Physics - Class 11 JEEM23Ph8342 Solution: A projectile is projected at from horizontal with initial…

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.

The initial horizontal component of the velocity is constant and is given by:

$v_{0x} = v_0 \cos\theta = 40\,\mathrm{ms}^{-1} \cos(30^{\circ}) = 40\,\mathrm{ms}^{-1} \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}\,\mathrm{ms}^{-1}$

The initial vertical component of the velocity is:

$v_{0y} = v_0 \sin\theta = 40\,\mathrm{ms}^{-1} \sin(30^{\circ}) = 40\,\mathrm{ms}^{-1} \cdot \frac{1}{2} = 20\,\mathrm{ms}^{-1}$

To find the vertical component of the velocity at t = 2 s, we use the equation:

$v_y = v_{0y} - gt = 20\,\mathrm{ms}^{-1} - (10\,\mathrm{ms}^{-2})(2\,\mathrm{s}) = 20\,\mathrm{ms}^{-1} - 20\,\mathrm{ms}^{-1} = 0\,\mathrm{ms}^{-1}$

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:

$\vec{v} = 20\sqrt{3}\,\mathrm{ms}^{-1} \hat{i} + 0\,\mathrm{ms}^{-1} \hat{j} = 20\sqrt{3}\,\mathrm{ms}^{-1}$

So the correct answer is: $20\sqrt{3}\,\mathrm{ms}^{-1}$