The Question is from neet-physics oscillations-waves , and it was created by the author ABDUL JILANI BASHA

Question

A particle is projected vertically upwards with an initial speed 'u'.  At a height 'h' above the ground, its speed is 'v'.  What is the maximum height reached by the particle?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Energy
	**Related Topics**: Kinematics, Potential Energy, Kinetic Energy
	**Prerequisites**: Understanding of kinetic and potential energy, equations of motion under constant acceleration (gravity).
## SOLUTION PATHWAY
**Given Information**:
	Initial velocity: u
	Velocity at height h: v
	Acceleration due to gravity: g (acting downwards)
	We are finding the maximum height reached by the particle.
**Method & Steps**:
	1.  Use the conservation of energy to relate the initial energy to the energy at height 'h'.
		Reasoning: The total mechanical energy (kinetic + potential) remains constant in the absence of non-conservative forces like friction.
		Formula/Rule: Initial Kinetic Energy = Kinetic Energy at Height h + Potential Energy at Height h
	2.  Express the kinetic energy at height h in terms of initial energy and the potential energy.
		Working: 1/2 * m * u^2 = 1/2 * m * v^2 + m * g * h
		Key Point: m is the mass of the particle, it cancels out.
	3.  Simplify the equation. u^2 = v^2 + 2 * g * h
	4.  Find the velocity at maximum height.  At maximum height, the velocity is zero. Let H be the maximum height.
		Reasoning: At maximum height, the particle momentarily stops before falling back down.
		Formula/Rule: 0 = v^2 - 2*g*(H-h), and using initial state 0 = u^2 - 2*g*H
	5.  Use conservation of energy to relate initial energy to energy at maximum height.
		Working: 1/2 * m * u^2 = m * g * H
	6.  From step 3 we have u^2 = v^2 + 2gh. Substituting 1/2 *m*u^2 = m*g*H gives us 1/2 *m*(v^2 + 2gh) = m*g*H.  Simplifying gives H = v^2/(2g) + h
		Result: H = v^2/(2g) + h
		Verification: The derived formula relates the maximum height to the given height and velocity at that height.
 ## OPTION ANALYSIS
[A] :x: This is the maximum height if the initial speed is u and the final speed is zero, but we are given that the speed is v at height h.
[B] :x: This is the maximum height if the initial speed is v and the final speed is zero, not valid for the given question.
[C] :x: 'h' is just the height where the particle has speed v, not the maximum height.
[D] ✓: This is the correct expression for the maximum height, derived from conservation of energy and kinematic equations.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly using the formula v^2 = u^2 + 2as for the entire motion rather than segments. [How to avoid]: Apply the formula only for sections with constant acceleration and carefully choose the initial and final states.
		2.[Error 2]: Forgetting that the velocity at the maximum height is zero. [How to avoid]: Always remember the kinematic conditions at critical points in motion like maximum height.
	**Quick Tips**:
		1.[Helpful shortcut]: When dealing with vertical motion problems, energy conservation is often a simpler approach than kinematic equations.
		2.[Memory aid]: Remember that the total mechanical energy (kinetic + potential) is conserved in the absence of non-conservative forces.

ABDUL JILANI BASHA · Answer

u²/2g

ABDUL JILANI BASHA · Answer

v²/2g

ABDUL JILANI BASHA · Answer

h

ABDUL JILANI BASHA · Answer

The correct Option is "v²/2g + h"

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