A projectile is launched from the surface of the earth with a very high speed u at an angle `theta` with vertical . What is its velocity when it is at the farthest distance from the earth surface . Given that the maximum height reached when it is launched vertically from the earth with a velocity v = `sqrt((GM)/(R))`

Options

A. `(u "cos" theta)/(2)`
B. `( u "sin" theta)/(2)`
C. `sqrt((GM)/(2R))`
D. `sqrt((GM)/(3R))`

Answer

To solve the problem step by step, we will analyze the projectile's motion and apply the principles of energy conservation and momentum conservation.

### Step 1: Understand the Problem
The projectile is launched from the Earth's surface with an initial speed \( u \) at an angle \( \theta \) with the vertical. We need to find its velocity when it reaches the farthest distance from the Earth's surface.

### Step 2: Maximum Height Calculation
When a projectile is launched vertically with a speed \( v = \sqrt{\frac{GM}{R}} \), it reaches a maximum height \( h \) given by the formula derived from energy conservation:
\[ h = \frac{v^2}{2g} \]
where \( g = \frac{GM}{R^2} \). Substituting \( v \):
\[ h = \frac{\left(\sqrt{\frac{GM}{R}}\right)^2}{2g} = \frac{\frac{GM}{R}}{2g} = \frac{R}{2} \]
Thus, the maximum height \( h \) when launched vertically is \( \frac{R}{2} \).

### Step 3: Work Done by Earth
The work done by the gravitational force when the projectile moves from radius \( R \) to \( R + h \) is given by:
\[ W = \int_{R}^{R+h} -\frac{GMm}{x^2} dx \]
Calculating this integral gives:
\[ W = -GMm \left[ \frac{1}{x} \right]_{R}^{R+h} = -GMm \left( \frac{1}{R+h} - \frac{1}{R} \right) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) \]

### Step 4: Energy Conservation
The work done by the Earth on the projectile is converted into kinetic energy:
\[ \frac{1}{2}mv^2 = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) \]
Cancelling \( m \) and rearranging gives:
\[ v^2 = 2GM \left( \frac{1}{R} - \frac{1}{R+h} \right) \]

### Step 5: Conservation of Momentum
Using conservation of momentum, we can relate the initial and final velocities:
\[ mv_0 \sin \theta = mv_f (R + h) \]
where \( v_0 = u \) is the initial velocity. Thus:
\[ u \sin \theta = v_f (R + h) \]

### Step 6: Substitute \( h \) and Solve for \( v_f \)
Substituting \( h \) into the equation:
\[ u \sin \theta = v_f \left(R + \frac{R}{2}\right) = v_f \left(\frac{3R}{2}\right) \]
Solving for \( v_f \):
\[ v_f = \frac{2u \sin \theta}{3} \]

### Conclusion
The final velocity of the projectile at the farthest distance from the Earth's surface is:
\[ v_f = \frac{2u \sin \theta}{3} \]