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Question

A particle is projected vertically upwards with a speed $u$ from the surface of the Earth. If the air resistance is proportional to the square of the velocity ($F = -kv^2$), what is the maximum height reached by the particle? (Assume $g$ is constant and $m$ is the mass of the particle)

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option B
## CONCEPT FRAMEWORK
	**Primary Concept**: Work-Energy Theorem with Non-Conservative Forces
	**Related Topics**:  Newton's Second Law, Integration, Air Resistance
	**Prerequisites**: Calculus (integration), understanding of forces and work, work-energy theorem.
## SOLUTION PATHWAY
**Given Information**:
	Initial velocity: $u$
	Air resistance: $F = -kv^2$
	Mass of particle: $m$
	Gravity: $g$ (constant)
	Final velocity at maximum height: $0$
**Method & Steps**:
	1.  **Apply Newton's Second Law:** While the particle is going up, the net force acting on it is the sum of gravity and air resistance, both acting downwards.  Therefore, $ma = -mg - kv^2$ where 'a' is the acceleration.
		Reasoning: Forces oppose the upward motion.
		Formula/Rule: $F = ma$
	2.  **Relate acceleration to velocity and position:** We know $a = \frac{dv}{dt}$. Also, $a = \frac{dv}{dx} \frac{dx}{dt} = v\frac{dv}{dx}$.
		Working:  Substituting $a=v\frac{dv}{dx}$ in the force equation from step 1, we get: $mv\frac{dv}{dx} = -mg - kv^2$.
		Key Point: This allows to relate velocity and displacement.
	3.  **Separate variables and integrate:** Rearranging the equation from step 2, we have $\frac{mv}{mg+kv^2} dv = -dx$. Integrate both sides with the initial condition x=0 when v=u and x=h when v=0.
		Working: Integrating we get $\int_{u}^{0} \frac{mv}{mg+kv^2} dv = \int_{0}^{h} -dx$.  Using substitution $w=mg+kv^2$, $dw=2kvdv$, we have  $\frac{m}{2k} \int_{mg+ku^2}^{mg} \frac{dw}{w} = -h$. This integrates to  $\frac{m}{2k} [\ln(mg) - \ln(mg+ku^2)]=-h$.  Or, $\frac{m}{2k} \ln(\frac{mg}{mg+ku^2}) = -h$. Using $\ln(a)-\ln(b)=\ln(a/b)$ and $\ln(1/x)=-\ln(x)$, we get  $h = \frac{m}{2k} \ln(1 + \frac{ku^2}{mg})$.
		Key Point: Integration gives the final height.
	4. **Final Answer**:
		Result: $h = \frac{m}{2k} \ln(1 + \frac{ku^2}{mg})$.
		Verification: The expression has the correct units and reduces to the expected form if air resistance is negligible (k->0).
 ## OPTION ANALYSIS
[A] :x: This is the maximum height without air resistance. It does not account for the air resistance
[B] ✓: This matches our derived result. It correctly accounts for the air resistance.
[C] :x: This has an incorrect factor (m/k instead of m/2k)
[D] :x: This has a negative sign in the logarithm, which is incorrect.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.  **Incorrect sign convention**: Ensure all forces are in the right direction.
		2.  **Incorrect integration limits**: Using the wrong limits leads to wrong results.
	**Quick Tips**:
		1. Use $a = v \frac{dv}{dx}$ to relate velocity and displacement when acceleration is velocity-dependent.
		2. Remember to properly separate variables before integration.

Mr. ANIL KUMAR.G · Answer

$\frac{u^2}{2g}$

Mr. ANIL KUMAR.G · Answer

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

Mr. ANIL KUMAR.G · Answer

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

Mr. ANIL KUMAR.G · Answer

$\frac{m}{2k} \ln(1 - \frac{ku^2}{mg})$

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