The Question is from neet-physics work-energy-power power, and it was created by the author P.NARASIMHA

Question

A particle of mass $m$ is projected vertically upwards with a velocity $v$.  If the air resistance is proportional to the square of the velocity, given by $kv^2$, where $k$ is a positive constant, what is the maximum height reached by the particle?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Work-Energy Theorem and Differential Equations
	**Related Topics**: Air resistance, Dynamics, Integration
	**Prerequisites**: Newton's second law, Basic calculus, integration techniques
## SOLUTION PATHWAY
**Given Information**:
	* Mass of the particle: $m$
	* Initial velocity: $v$
	* Air resistance force: $kv^2$ (acting downwards during upward motion)
	* Acceleration due to gravity: $g$
	* We are looking for the maximum height reached by the particle.
**Method & Steps**:
	1.  Set up the equation of motion using Newton's second law. The net force acting on the particle during its upward motion is the sum of the gravitational force ($-mg$) and the air resistance force ($-kv^2$).
		Reasoning: Newton's second law states that the net force on an object is equal to its mass times its acceleration.
		Formula/Rule: $F_{net} = ma$
	2.  Write the differential equation for the particle's motion. The acceleration is the time derivative of velocity, which in turn is the time derivative of position. However, it's more convenient to relate acceleration to velocity directly.
		Working:  $ma = -mg - kv^2$, so $a = \frac{dv}{dt} = -g - \frac{k}{m}v^2$. Since $a = v\frac{dv}{dy}$, we have $v\frac{dv}{dy} = -g - \frac{k}{m}v^2$.
		Key Point: We use $a = v\frac{dv}{dy}$ to relate acceleration with position.
	3.  Separate variables and integrate to find the relationship between velocity and position. Rearranging the equation, we get $\frac{v dv}{g + (k/m)v^2} = -dy$. Integrating both sides, we get  $\int_{v}^{0} \frac{v dv}{g + (k/m)v^2} = -\int_{0}^{h} dy$, where $h$ is the maximum height.
		Working: Let $u = g + (k/m)v^2$, then $du = (2k/m)v dv$, so $v dv = \frac{m}{2k}du$. The integral becomes $\frac{m}{2k}\int \frac{du}{u} = \frac{m}{2k} \ln(u)$.  Evaluating from $v$ to 0, we get $\frac{m}{2k}[\ln(g) - \ln(g + \frac{k}{m}v^2)] = -h$, so $h = \frac{m}{2k} \ln(\frac{g + (k/m)v^2}{g}) = \frac{m}{2k}\ln(1 + \frac{kv^2}{mg})$.
		Result: $h = \frac{m}{2k} \ln(1 + \frac{kv^2}{mg})$
		Verification: The result is consistent with the dimensions of the problem; each term in the argument of the logarithm is dimensionless, and the overall result has dimensions of length.
 ## OPTION ANALYSIS
[A] :x: This option has a factor of $\frac{m}{k}$ instead of $\frac{m}{2k}$ and the logarithm term is also different. 
[B] :x: This option has the correct $\frac{m}{2k}$ factor but the logarithm term is incorrect.
[C] ✓: This option matches the derived expression for maximum height.
[D] :x: This option has an incorrect factor of $\frac{m}{k}$ and an incorrect coefficient in the logarithm term. 
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Incorrectly applying the chain rule**: When relating acceleration to position, use $a=v\frac{dv}{dy}$, not $a = \frac{dv}{dy}$.
		2. **Sign errors**: Be careful with the signs of the forces involved. Air resistance acts opposite to the direction of motion.
	**Quick Tips**:
		1. **Use a substitution**: When integrating, using a substitution like $u = g + \frac{k}{m}v^2$ can simplify the integral.
		2. **Dimensional analysis**: Check if the final answer has the correct dimensions. In this case, the height must have the dimensions of length.

P.NARASIMHA · Answer

$\frac{m}{k} \ln(1 + \frac{kv_0^2}{mg})$

P.NARASIMHA · Answer

$\frac{m}{2k} \ln(1 + \frac{kv_0^2}{2mg})$

P.NARASIMHA · Answer

The correct Option is "$\frac{m}{2k} \ln(1 + \frac{kv_0^2}{mg})$"

P.NARASIMHA · Answer

$\frac{m}{k} \ln(1 + \frac{2kv_0^2}{mg})$

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