Import Question JSON

Current Question (ID: 9632)

Question:

$\text{A projectile is projected from the ground with the velocity } v_0 \text{ at an angle } \theta \text{ with the horizontal. What is the vertical component of the velocity of the projectile when its vertical displacement is equal to half of the maximum height attained?}$

Options:

1. $\sqrt{3}v_0 \cos \theta$
2. $\frac{v_0}{\sqrt{2}} \sin \theta$
3. $\frac{v_0}{\sqrt{2}} \cos \theta$
4. $\sqrt{5}v_0$

Solution:

\text{Hint: } H = \frac{u^2 \sin^2 \theta}{2g} \text{Step: Find the vertical component of the velocity of the projectile.} \text{The maximum height attained by the projectile is given by:} H = \frac{u^2 \sin^2 \theta}{2g} \text{Using the kinematic equation: } v_y^2 = u_y^2 - 2gy \text{At maximum height, the vertical displacement } y = H \text{ and } u_y = u \sin \theta v_y^2 = u^2 \sin^2 \theta - 2g \times \frac{u^2 \sin^2 \theta}{2g} v_y^2 = u^2 \sin^2 \theta - u^2 \sin^2 \theta = 0 \text{At maximum height, } v_y = 0 \text{At half the maximum height, } y = \frac{H}{2} = \frac{u^2 \sin^2 \theta}{4g} v_y^2 = u^2 \sin^2 \theta - 2g \times \frac{u^2 \sin^2 \theta}{4g} = u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{2} = \frac{u^2 \sin^2 \theta}{2} \text{Therefore, } v_y = \frac{u \sin \theta}{\sqrt{2}} \text{Hence, option (2) is the correct answer.}

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}