Two projectiles are launched from the same point with the same speed $u$. One is launched at an angle $\theta$ and the other at $(90^\circ - \theta)$ with the horizontal. What is the minimum distance between these two projectiles during their flight?

## CORE INFORMATION\n✓ **Answer**: A\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Projectile Motion\n\t**Related Topics**: Kinematics, Relative Motion\n\t**Prerequisites**: Understanding of projectile motion equations, concept of relative velocity\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Two projectiles launched with same speed $u$\n\t* Launch angles: $\theta$ and $(90^\circ - \theta)$\n\t* Launched from the same point\n**Method & Steps**:\n\t1.\tAnalyze the motion of the projectiles in terms of components.\n\t\tReasoning: We can represent the velocity in terms of horizontal and vertical components. Both projectiles have the same initial velocity magnitude. Let's consider the horizontal and vertical components of the initial velocity.\n\t\tFormula/Rule: $v_x = u \cos(\alpha)$, $v_y = u \sin(\alpha)$, where $\alpha$ is the launch angle.\n\t2.\tDetermine the relative velocities of the two projectiles.\n\t\tWorking: Projectile 1 has velocity components $u\cos(\theta)$ and $u\sin(\theta)$. Projectile 2 has velocity components $u\cos(90^\circ - \theta) = u\sin(\theta)$ and $u\sin(90^\circ - \theta) = u\cos(\theta)$. The relative horizontal velocity is $u\cos(\theta) - u\sin(\theta)$ and the relative vertical velocity is $u\sin(\theta) - u\cos(\theta)$.\n\t\tKey Point: Notice that the horizontal component of projectile 1 is the vertical component of projectile 2, and vice versa.\n\t3.\tAnalyze the relative motion. \n\t\tReasoning: Let's examine the relative velocity. The relative horizontal velocity is $u\cos\theta - u\sin\theta$ and the relative vertical velocity is $u\sin\theta - u\cos\theta$. The relative velocity vector is orthogonal to the vector joining the projectiles.\n\t4.\tDetermine the minimum distance.\n\t\tReasoning: Since the projectiles start from the same point and have velocities that are just interchanged between horizontal and vertical components, they will always be at a distance equal to the relative velocity multiplied by time. The relative velocity vector is constant and the relative position vector is always perpendicular to it. This means the distance is changing linearly with time.\n\t\tResult: The minimum distance is 0 because they start from the same point and their relative motion is such that they will always be separated by a distance that is directly proportional to time.\n\t\tVerification: Because the initial position is the same, the minimum distance must be zero, which occurs at the start of the motion.\n ## OPTION ANALYSIS\n[A] ✓: The minimum distance is 0 because the projectiles start at the same point.\n[B] :x: This is the formula for the maximum height of a projectile launched at 45 degrees, not the minimum distance between two projectiles.\n[C] :x: This is not a standard formula related to relative projectile motion.\n[D] :x: This is the range of a projectile launched at a 45 degree angle, not the minimum distance.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Confusing the horizontal and vertical components of the velocities. Make sure you are using the correct trigonometric functions.\n\t\t2.[Error 2]: Assuming that the distance is not zero at the beginning of the motion. Remember that the projectiles start at the same point.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: Visualize the motion: the projectiles are essentially swapping their horizontal and vertical components, making the relative motion linear from the start.\n\t\t2.[Memory aid]: The relative motion is always from the same point of origin, so the minimum distance is always zero.

NEET Chemistry MCQs

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NEET Questions / Chemistry

A 20 L vessel contains 0.1 mole of $O_2$ at 300 K. $O_3$ is prepared in the vessel by subjecting $O_2$ to silent electric discharge till the volume is reduced to 15 L at 300K. What is the number of moles of $O_3$ formed?

0.05

0.025

0.075

0.01

Question Tags

📚 Subject:Chemistry

📑 Chapter:Some Basic Concepts in Chemistry

🎯Topic:Mole Concept

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :4 mins

🔄Ncert Concepts :Mole Concept, Avogadro's Law, Stoichiometry

A compound containing only carbon, hydrogen, and oxygen has a molar mass of approximately 90 g/mol. Combustion analysis of 0.405 g of the compound produces 0.54 g $CO_2$ and 0.27 g $H_2O$ . If the compound's empirical formula is the same as its molecular formula, what is the number of oxygen atoms in one molecule of the compound?

Question Tags

📚 Subject:Chemistry

📑 Chapter:Some Basic Concepts in Chemistry

🎯Topic:Emprirical Formula & Molecular Formula

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :4 mins

🔄Ncert Concepts :Combustion Analysis, Empirical Formula, Molecular Formula, Stoichiometry

A projectile is launched with initial speed $u$ at an angle $heta$ from the horizontal. At the highest point, it explodes into three fragments of equal mass. One fragment falls vertically downwards, the second fragment continues along the original path with speed $v$ . Find the speed of the third fragment immediately after the explosion.

3u cos θ - v

|3u cos θ - v|

v - 3u cos θ

-(3u cos θ + v)

📚 Subject:Chemistry

📑 Chapter:Chemical Bonding and Molecular Structure

🎯Topic:Dipole Moment

🎓 Ncert Level:Analyzing

🎯Difficulty :Hard

⚡ Time Required :4 mins

🔄Ncert Concepts :Dipole moment, Bond angle, Percent ionic character, Molecular geometry

20 mL of a solution containing a mixture of $NaOH$ and $Na_2CO_3$ requires 10 mL of 0.5 N HCl for neutralization using phenolphthalein as an indicator. A further addition of 5 mL of the same HCl is required to reach the methyl orange endpoint. Calculate the equivalent weight of the mixture.

66.67

106

Question Tags

📚 Subject:Chemistry

📑 Chapter:Some Basic Concepts in Chemistry

🎯Topic:Equivalent Weight

🎓 Ncert Level:Analyzing

🎯Difficulty :Hard

⚡ Time Required :4 mins

🔄Ncert Concepts :Equivalent Weight, Double Indicator Titration, Stoichiometry

10.

An element X forms a chloride with the formula XCl $_3$ and another chloride with the formula XCl $_5$ . 0.1 moles of XCl $_3$ requires 60 mL of 1 N NaOH for complete hydrolysis, while 0.1 moles of XCl $_5$ requires 100 mL of 1 N NaOH. What is the equivalent weight of the element X in XCl $_3$ and XCl $_5$ respectively?