A body is thrown from the surface of the earth with velocity u m/s. The maximum height in metre above the surface of the earth upto which it will reach is (where, R = radish of earth, g=acceleration due to gravity)

Options

A. `(u^(2)R)/(2gR-u^(2))`
B. `(2u^(2)R)/(gR-u^(2))`
C. `(u^(2)R^(2))/(2gR^(2)-u^(2))`
D. `(u^(2)R)/(gR-u^(2))`

Answer

To solve the problem of finding the maximum height \( h \) above the surface of the Earth that a body reaches when thrown with an initial velocity \( u \), we can use the principle of conservation of mechanical energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Energy Conservation Principle
The total mechanical energy (kinetic + potential) at the surface of the Earth will equal the total mechanical energy at the maximum height. At maximum height, the kinetic energy will be zero because the body momentarily stops before falling back down.

### Step 2: Write the Initial and Final Energy Expressions
- **Initial Energy (at the surface)**:
\[ E_i = \text{Kinetic Energy} + \text{Potential Energy} = \frac{1}{2} m u^2 - \frac{GMm}{R} \]
where \( m \) is the mass of the body, \( G \) is the gravitational constant, and \( R \) is the radius of the Earth.

- **Final Energy (at maximum height)**:
\[ E_f = \text{Kinetic Energy} + \text{Potential Energy} = 0 + \left(-\frac{GMm}{R+h}\right) \]

### Step 3: Set Initial Energy Equal to Final Energy
Using the conservation of energy:
\[ \frac{1}{2} m u^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

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

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{1}{2} u^2 = \frac{GM}{R} - \frac{GM}{R+h} \]

### Step 6: Finding a Common Denominator
The right-hand side can be simplified:
\[ \frac{1}{2} u^2 = GM \left( \frac{(R+h) - R}{R(R+h)} \right) = \frac{GMh}{R(R+h)} \]

### Step 7: Solve for \( h \)
Cross-multiplying gives:
\[ \frac{1}{2} u^2 R(R+h) = GMh \]
Expanding and rearranging:
\[ \frac{1}{2} u^2 R^2 + \frac{1}{2} u^2 Rh - GMh = 0 \]
This is a quadratic equation in \( h \):
\[ \left(\frac{1}{2} u^2 R - GM\right)h + \frac{1}{2} u^2 R^2 = 0 \]

### Step 8: Use the Quadratic Formula
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = \frac{1}{2} u^2 R - GM \), \( b = 0 \), and \( c = \frac{1}{2} u^2 R^2 \).

### Step 9: Substitute Values and Simplify
After solving, we find:
\[ h = \frac{u^2 R}{2g} - \frac{u^2 R}{GM} \]
Where \( g = \frac{GM}{R^2} \).

### Final Result
Thus, the maximum height \( h \) above the Earth's surface is given by:
\[ h = \frac{u^2 R}{2g} - \frac{u^2 R}{GM} \]