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_(E))/(1-k^(2))`
B. `(R_(E))/(k^(2))`
C. `(1-k^(2))/(R_(E))`
D. `(k^(2))/(R_(E))`

Explanation

To solve the problem of determining the maximum height a projectile will reach when fired vertically upwards from the surface of the Earth with a velocity \( K v_e \) (where \( v_e \) is the escape velocity and \( K < 1 \)), we can follow these steps:

### Step 1: Understand the Initial Conditions
The projectile is fired from the surface of the Earth with an initial velocity \( K v_e \). 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
The initial kinetic energy \( KE_i \) of the projectile can be expressed as:
\[ KE_i = \frac{1}{2} m (K v_e)^2 = \frac{1}{2} m K^2 v_e^2 \]

### Step 3: Calculate Initial Potential Energy
The initial potential energy \( PE_i \) at the surface of the Earth is given by:
\[ PE_i = -\frac{GMm}{R} \]

### Step 4: Set Up Conservation of Energy
At the maximum height \( h \), the final kinetic energy \( KE_f \) is zero (since the projectile momentarily stops before falling back). The potential energy \( PE_f \) at height \( h \) is:
\[ PE_f = -\frac{GMm}{R + h} \]
According to the conservation of energy:
\[ KE_i + PE_i = KE_f + PE_f \]
Substituting the values we have:
\[ \frac{1}{2} m K^2 v_e^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R + h} \]

### Step 5: Simplify the Equation
We can cancel \( m \) from all terms (assuming \( m \neq 0 \)):
\[ \frac{1}{2} K^2 v_e^2 - \frac{GM}{R} = -\frac{GM}{R + h} \]
Rearranging gives:
\[ \frac{1}{2} K^2 v_e^2 = \frac{GM}{R} - \frac{GM}{R + h} \]

### Step 6: Substitute for Escape Velocity
Substituting \( v_e^2 = \frac{2GM}{R} \) into the equation:
\[ \frac{1}{2} K^2 \left(\frac{2GM}{R}\right) = \frac{GM}{R} - \frac{GM}{R + h} \]
This simplifies to:
\[ K^2 \frac{GM}{R} = \frac{GM}{R} - \frac{GM}{R + h} \]

### Step 7: Factor Out Common Terms
Factoring out \( GM \):
\[ K^2 \frac{1}{R} = \frac{1}{R} - \frac{1}{R + h} \]
This can be rewritten as:
\[ K^2 = 1 - \frac{R}{R + h} \]

### Step 8: Solve for \( h \)
Rearranging gives:
\[ K^2 = \frac{h}{R + h} \]
Cross-multiplying yields:
\[ K^2 (R + h) = h \]
This simplifies to:
\[ K^2 R + K^2 h = h \]
Rearranging gives:
\[ h - K^2 h = K^2 R \]
Factoring out \( h \):
\[ h(1 - K^2) = K^2 R \]
Thus,
\[ h = \frac{K^2 R}{1 - K^2} \]

### Step 9: Calculate Total Height from the Center of the Earth
The total height \( H \) from the center of the Earth is:
\[ H = R + h = R + \frac{K^2 R}{1 - K^2} = R \left(1 + \frac{K^2}{1 - K^2}\right) \]
This simplifies to:
\[ H = R \left(\frac{1 - K^2 + K^2}{1 - K^2}\right) = \frac{R}{1 - K^2} \]

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