A projectile is fired vertically upwards from the surface of the earth with a velocity `Kv_(e)`, where `v_(e)` is the escape velocity and `Klt1`.If `R` is the radius of the earth, the maximum height to which it will rise measured from the centre of the earth will be (neglect air resistance)

Options

A. `R/(K^(2)-1)`
B. `R/(1-K^(2))`
C. `R(1-K^(2))`
D. `R/(1+K^(2))`

Explanation

To solve the problem of finding the maximum height to which a projectile will rise when fired vertically upwards with a velocity \( Kv_e \) (where \( v_e \) is the escape velocity and \( K < 1 \)), we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The projectile is fired with an initial velocity \( v_0 = Kv_e \). The escape velocity \( v_e \) is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 2: Calculate Initial Kinetic Energy and Potential Energy
At the surface of the Earth, the initial kinetic energy \( KE_i \) and potential energy \( PE_i \) are:
\[ KE_i = \frac{1}{2} m (Kv_e)^2 = \frac{1}{2} m K^2 v_e^2 \]
Substituting \( v_e^2 \):
\[ KE_i = \frac{1}{2} m K^2 \left(\frac{2GM}{R}\right) = \frac{m K^2 GM}{R} \]
The potential energy at the surface \( PE_i \) is:
\[ PE_i = -\frac{GMm}{R} \]

### Step 3: Total Mechanical Energy at the Surface
The total mechanical energy \( E_i \) at the surface is:
\[ E_i = KE_i + PE_i = \frac{m K^2 GM}{R} - \frac{GMm}{R} = \frac{GMm}{R} (K^2 - 1) \]

### Step 4: Energy at Maximum Height
At the maximum height \( H \), the kinetic energy \( KE_f \) will be zero, and the potential energy \( PE_f \) is given by:
\[ PE_f = -\frac{GMm}{H} \]
The total mechanical energy at maximum height \( E_f \) is:
\[ E_f = KE_f + PE_f = 0 - \frac{GMm}{H} = -\frac{GMm}{H} \]

### Step 5: Apply Conservation of Energy
According to the conservation of mechanical energy:
\[ E_i = E_f \]
Substituting the expressions for \( E_i \) and \( E_f \):
\[ \frac{GMm}{R} (K^2 - 1) = -\frac{GMm}{H} \]

### Step 6: Rearranging the Equation
Cancelling \( GMm \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{K^2 - 1}{R} = -\frac{1}{H} \]
Rearranging gives:
\[ \frac{1}{H} = \frac{1 - K^2}{R} \]
Thus,
\[ H = \frac{R}{1 - K^2} \]

### Conclusion
The maximum height \( H \) to which the projectile will rise, measured from the center of the Earth, is:
\[ H = \frac{R}{1 - K^2} \]