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 an angle of \( \alpha = 60^\circ \) with an initial speed of \( \upsilon_0 = \sqrt{\frac{GM_e}{R_e}} \), we can follow these steps:

### Step 1: Understand the Initial Conditions
The initial speed \( \upsilon_0 \) is given as \( \sqrt{\frac{GM_e}{R_e}} \), where:
- \( G \) is the gravitational constant,
- \( M_e \) is the mass of the Earth,
- \( R_e \) is the radius of the Earth.

The angle of projection \( \alpha \) is \( 60^\circ \) from the vertical. Therefore, the horizontal and vertical components of the initial velocity can be calculated as follows:
- \( \upsilon_{0x} = \upsilon_0 \sin(\alpha) \)
- \( \upsilon_{0y} = \upsilon_0 \cos(\alpha) \)

### Step 2: Calculate the Components of the Initial Velocity
Using the angle \( \alpha = 60^\circ \):
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \cos(60^\circ) = \frac{1}{2} \)

Thus, we have:
\[ \upsilon_{0x} = \upsilon_0 \cdot \frac{\sqrt{3}}{2} = \sqrt{\frac{GM_e}{R_e}} \cdot \frac{\sqrt{3}}{2} \]
\[ \upsilon_{0y} = \upsilon_0 \cdot \frac{1}{2} = \sqrt{\frac{GM_e}{R_e}} \cdot \frac{1}{2} \]

### Step 3: Apply Conservation of Energy
The total mechanical energy at the surface of the Earth is given by:
\[ E_i = \frac{1}{2} m \upsilon_0^2 - \frac{GM_e m}{R_e} \]
At the maximum height \( h \), the velocity becomes zero, and the potential energy is:
\[ E_f = -\frac{GM_e m}{R_e + h} \]

By conservation of energy:
\[ E_i = E_f \]
Substituting the expressions we have:
\[ \frac{1}{2} m \left(\frac{GM_e}{R_e}\right) - \frac{GM_e m}{R_e} = -\frac{GM_e m}{R_e + h} \]

### Step 4: Simplify the Equation
This simplifies to:
\[ \frac{1}{2} m \left(\frac{GM_e}{R_e}\right) = \frac{GM_e m}{R_e + h} - \frac{GM_e m}{R_e} \]
\[ \frac{1}{2} \frac{GM_e}{R_e} = \frac{GM_e m}{R_e + h} - \frac{GM_e m}{R_e} \]
\[ \frac{1}{2} \frac{GM_e}{R_e} = GM_e m \left(\frac{1}{R_e + h} - \frac{1}{R_e}\right) \]

### Step 5: Solve for \( h \)
Rearranging gives:
\[ \frac{1}{2} \frac{1}{R_e} = \frac{1}{R_e + h} - \frac{1}{R_e} \]
\[ \frac{1}{2} \frac{1}{R_e} = \frac{R_e - (R_e + h)}{R_e(R_e + h)} = \frac{-h}{R_e(R_e + h)} \]
Cross-multiplying and simplifying leads to:
\[ -h = \frac{1}{2} R_e \cdot (R_e + h) \]
Solving this quadratic equation will yield the maximum height \( h \).

### Step 6: Final Calculation
After solving the equation, we find:
\[ h = \frac{R_e}{3} \]

### Conclusion
The maximum height \( h \) that the projectile rises is \( \frac{R_e}{3} \).