A missile is launched at an angle of `60^(@)` to the vertical with a velocity `sqrt( 0.75gR)` from the surface of the earth ( R is the radius of the earth). Find the maximum height from the surface of earth. (Neglect air resistance and rotating of earth).

Answer

To find the maximum height from the surface of the Earth for a missile launched at an angle of \(60^\circ\) to the vertical with a velocity of \(\sqrt{0.75gR}\), we can follow these steps:

### Step 1: Determine the initial velocity components
The missile is launched with a speed \(u = \sqrt{0.75gR}\) at an angle of \(60^\circ\) to the vertical. We can break this velocity into its vertical and horizontal components.

- The vertical component of the initial velocity \(u_y\) is given by:
\[ u_y = u \cos(60^\circ) = \sqrt{0.75gR} \cdot \frac{1}{2} = \frac{\sqrt{0.75gR}}{2} \]

- The horizontal component of the initial velocity \(u_x\) is given by:
\[ u_x = u \sin(60^\circ) = \sqrt{0.75gR} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3 \cdot 0.75gR}}{2} \]

### Step 2: Apply conservation of mechanical energy
At the maximum height \(h\), the vertical component of the velocity becomes zero. We can use the conservation of mechanical energy to find the maximum height.

The total mechanical energy at the launch point (point A) is:
\[ E_A = K.E_A + P.E_A = \frac{1}{2} m u^2 - \frac{GMm}{R} \]

At the maximum height (point B), the total mechanical energy is:
\[ E_B = K.E_B + P.E_B = 0 + \left(-\frac{GMm}{R + h}\right) \]

Setting \(E_A = E_B\):
\[ \frac{1}{2} m u^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

### Step 3: Substitute \(u\) and simplify
Substituting \(u = \sqrt{0.75gR}\):
\[ \frac{1}{2} m (0.75gR) - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

Canceling \(m\) from both sides:
\[ \frac{0.75gR}{2} - \frac{GM}{R} = -\frac{GM}{R + h} \]

### Step 4: Use \(g = \frac{GM}{R^2}\)
Substituting \(g\) into the equation:
\[ \frac{0.75 \cdot \frac{GM}{R^2} R}{2} - \frac{GM}{R} = -\frac{GM}{R + h} \]

This simplifies to:
\[ \frac{0.375 GM}{R} - \frac{GM}{R} = -\frac{GM}{R + h} \]
\[ -\frac{0.625 GM}{R} = -\frac{GM}{R + h} \]

### Step 5: Cross-multiply and solve for \(h\)
Cross-multiplying gives:
\[ 0.625 GM (R + h) = GM R \]
Dividing by \(GM\) (assuming \(GM \neq 0\)):
\[ 0.625(R + h) = R \]
\[ 0.625R + 0.625h = R \]
\[ 0.625h = R - 0.625R \]
\[ 0.625h = 0.375R \]
\[ h = \frac{0.375R}{0.625} = 0.6R \]

### Conclusion
The maximum height \(h\) from the surface of the Earth is:
\[ h = 0.6R \]