JEE MAIN - Physics (2023 - 11th April Evening Shift - No. 12)
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 $$g=10 \mathrm{~m} / \mathrm{s}^{2}$$ )
$$20 \sqrt{3} \mathrm{~ms}^{-1}$$
Zero
$$20 \mathrm{~ms}^{-1}$$
$$40 \sqrt{3} \mathrm{~ms}^{-1}$$
Explanation
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: $$ 20\sqrt{3}\,\mathrm{ms}^{-1} $$
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: $$ 20\sqrt{3}\,\mathrm{ms}^{-1} $$
Comments (0)
