Solution to the Problem

Deriving the Energy Equation for the Motion

The motion of the particle can be analyzed using the conservation of mechanical energy. The total energy (sum of kinetic energy and potential energy) remains constant throughout the motion, neglecting air resistance.

1. At ground level (initial position):

   • Kinetic Energy (KE) = $\frac{1}{2} m u^2$
   • Potential Energy (PE) = $0$ (taking the potential energy at the ground as zero)
   • Total Energy, $E_{\text{total}}$ = $\frac{1}{2} m u^2$

2. At a height $h$ during motion:

   • Kinetic Energy = $\frac{1}{2} m v^2$ (where $v$ is the speed at height $h$)
   • Potential Energy = $m g h$
   • Total Energy = $\frac{1}{2} m v^2 + m g h$

Using the conservation of total energy: $\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + m g h$

Simplify: $\frac{1}{2} u^2 = \frac{1}{2} v^2 + g h$

This is the energy equation for the motion under gravity.

Determining the Maximum Height

At the maximum height, the particle comes to rest momentarily, so $v = 0$.

Substitute $v = 0$ in the energy equation: $\frac{1}{2} u^2 = g h_{\text{max}}$

Solve for $h_{\text{max}}$: $h_{\text{max}} = \frac{u^2}{2 g}$

Thus, the maximum height reached by the particle is: $h_{\text{max}} = \frac{u^2}{2 g}$

Determining the Speed When the Particle Falls Halfway Back to Earth

At halfway back to earth, the height of the particle is: $h = \frac{h_{\text{max}}}{2} = \frac{u^2}{4 g}$

Using the energy equation: $\frac{1}{2} u^2 = \frac{1}{2} v^2 + g h$

Substitute $h = \frac{u^2}{4 g}$: $\frac{1}{2} u^2 = \frac{1}{2} v^2 + g \left(\frac{u^2}{4 g}\right)$

Simplify: $\frac{1}{2} u^2 = \frac{1}{2} v^2 + \frac{u^2}{4}$

Rearrange for $v^2$: $\frac{1}{2} v^2 = \frac{1}{2} u^2 - \frac{u^2}{4}$

$\frac{1}{2} v^2 = \frac{2 u^2}{4} - \frac{u^2}{4}$

$\frac{1}{2} v^2 = \frac{u^2}{4}$

$v^2 = \frac{u^2}{2}$

Solve for $v$: $v = \frac{u}{\sqrt{2}}$

Thus, the speed of the particle when it has fallen halfway back is: $v = \frac{u}{\sqrt{2}}$

Final Results

1. Energy Equation: $\frac{1}{2} u^2 = \frac{1}{2} v^2 + g h$

2. Maximum Height: $h_{\text{max}} = \frac{u^2}{2 g}$

3. Speed at Halfway Point: $v = \frac{u}{\sqrt{2}}$

Would you like further clarification or more detailed derivations?

Related Questions

1. How does the principle of energy conservation apply in this problem?
2. What assumptions are made about external forces in this derivation?
3. Can we derive the time taken to reach maximum height using this information?
4. How would the solution change if air resistance were considered?
5. What is the physical significance of the particle's kinetic energy and potential energy at the halfway point?

Tip

When solving motion problems, always clearly define the reference point for potential energy to avoid confusion in energy calculations.