A projectile of mass `m` is fired from the surface of the earth at an angle `alpha = 60^(@)` from the vertical. The initial speed `upsilon_(0)` is equal to `sqrt((GM_(e))/(R_(e))`. How high does the projectile rise ? Neglect air resistance and the earth's rotation.

Explanation

To solve the problem of how high the projectile rises when fired from the surface of the Earth at a given angle, we can follow these steps:

### Step 1: Understand the Initial Conditions
The projectile is fired with an initial speed \( v_0 = \sqrt{\frac{GM_e}{R_e}} \) at an angle \( \alpha = 60^\circ \) from the vertical. Here, \( G \) is the gravitational constant, \( M_e \) is the mass of the Earth, and \( R_e \) is the radius of the Earth.

### Step 2: Resolve the Initial Velocity
Since the projectile is fired at an angle of \( 60^\circ \) from the vertical, we can resolve the initial velocity into two components:
- Vertical component: \( v_{0y} = v_0 \cos(60^\circ) = v_0 \cdot \frac{1}{2} = \frac{1}{2} v_0 \)
- Horizontal component: \( v_{0x} = v_0 \sin(60^\circ) = v_0 \cdot \frac{\sqrt{3}}{2} \)

### Step 3: Apply Conservation of Angular Momentum
At the maximum height, the vertical component of the velocity becomes zero. The angular momentum at the surface can be expressed as:
\[ L_i = m v_0 R_e \sin(60^\circ) = m v_0 R_e \cdot \frac{\sqrt{3}}{2} \]
At maximum height \( r_{max} \), the angular momentum is:
\[ L_f = m v_{max} r_{max} \]
Setting these equal gives:
\[ m v_0 R_e \cdot \frac{\sqrt{3}}{2} = m v_{max} r_{max} \]
Cancelling \( m \) and rearranging gives:
\[ v_{max} = \frac{v_0 R_e \cdot \frac{\sqrt{3}}{2}}{r_{max}} \]

### Step 4: Apply Conservation of Mechanical Energy
The total mechanical energy at the surface is equal to the total mechanical energy at the maximum height:
\[ \frac{1}{2} m v_0^2 - \frac{GM_e m}{R_e} = \frac{1}{2} m v_{max}^2 - \frac{GM_e m}{r_{max}} \]
Substituting \( v_0^2 = \frac{GM_e}{R_e} \) into the equation gives:
\[ \frac{1}{2} m \left(\frac{GM_e}{R_e}\right) - \frac{GM_e m}{R_e} = \frac{1}{2} m v_{max}^2 - \frac{GM_e m}{r_{max}} \]
Simplifying this leads to:
\[ -\frac{1}{2} \frac{GM_e m}{R_e} = \frac{1}{2} m v_{max}^2 - \frac{GM_e m}{r_{max}} \]

### Step 5: Solve for Maximum Height
From the conservation of energy equation, we can isolate \( r_{max} \) and solve for it. After simplification, we find:
\[ r_{max} = \frac{3}{2} R_e \]
The height \( h_{max} \) that the projectile rises above the Earth's surface is:
\[ h_{max} = r_{max} - R_e = \frac{3}{2} R_e - R_e = \frac{1}{2} R_e \]

### Final Answer
Thus, the maximum height \( h_{max} \) that the projectile rises is:
\[ h_{max} = \frac{R_e}{2} \]

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