In the previous problem, the minimum speed with which the body must be thrown from the surface of the earth so as to reach a height of `R//4` is

Options

A. `sqrt((gR)/2)`
B. `sqrt(gR)`
C. `sqrt((gR)/5)`
D. `sqrt((2gR)/5)`

Explanation

To solve the problem of finding the minimum speed with which a body must be thrown from the surface of the Earth to reach a height of \( \frac{R}{4} \), we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
When the body is thrown upwards, its mechanical energy (kinetic + potential) at the surface of the Earth will be equal to its mechanical energy at the height \( \frac{R}{4} \). At the maximum height, the kinetic energy will be zero, and only potential energy will be present.

### Step 2: Write the Expression for Mechanical Energy at Two Points
1. **At the surface (Point 1)**:
- Kinetic Energy (KE) = \( \frac{1}{2} mv^2 \)
- Potential Energy (PE) = \( -\frac{GMm}{R} \) (where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth)

Total Energy at Point 1:
\[ E_1 = \frac{1}{2} mv^2 - \frac{GMm}{R} \]

2. **At the height \( \frac{R}{4} \) (Point 2)**:
- Kinetic Energy (KE) = 0 (at maximum height)
- Potential Energy (PE) = \( -\frac{GMm}{\frac{5R}{4}} = -\frac{4GMm}{5R} \)

Total Energy at Point 2:
\[ E_2 = 0 - \frac{4GMm}{5R} = -\frac{4GMm}{5R} \]

### Step 3: Set the Energies Equal
According to the conservation of mechanical energy:
\[ E_1 = E_2 \]
Substituting the expressions we derived:
\[ \frac{1}{2} mv^2 - \frac{GMm}{R} = -\frac{4GMm}{5R} \]

### Step 4: Simplify the Equation
1. Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 - \frac{GM}{R} = -\frac{4GM}{5R} \]

2. Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{GM}{R} - \frac{4GM}{5R} \]

3. Combine the terms on the right:
\[ \frac{1}{2} v^2 = \frac{5GM}{5R} - \frac{4GM}{5R} = \frac{1GM}{5R} \]

### Step 5: Solve for \( v^2 \)
Multiply both sides by 2:
\[ v^2 = \frac{2GM}{5R} \]

### Step 6: Take the Square Root
\[ v = \sqrt{\frac{2GM}{5R}} \]

### Step 7: Substitute \( g \)
Using the relation \( g = \frac{GM}{R^2} \), we can express \( GM \) as \( gR \):
\[ v = \sqrt{\frac{2gR}{5}} \]

### Final Result
Thus, the minimum speed with which the body must be thrown from the surface of the Earth to reach a height of \( \frac{R}{4} \) is:
\[ v = \sqrt{\frac{2gR}{5}} \]