Show that if a body be projected vertically upward from the surface of the earth so as to reach a height `nR` above the surface
(i) the increase in its potential energy is `((n)/(n + 1)) mgR`,
(ii) the velocity with it must be projected is `sqrt((2ngR)/(n + 1))`, where `r` is the radius of the earth and `m` the mass of body.

Explanation

To solve the problem, we will break it down into two parts:

1. **Finding the increase in potential energy** when a body is projected to a height of \( nR \) above the surface of the Earth.
2. **Finding the velocity** with which the body must be projected to reach that height.

### Part (i): Increase in Potential Energy

1. **Initial Potential Energy (U_initial)**:
The potential energy at the surface of the Earth is given by the formula:
\[ U_{\text{initial}} = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the body, and \( R \) is the radius of the Earth.

2. **Final Potential Energy (U_final)**:
When the body reaches a height of \( nR \), the total distance from the center of the Earth becomes \( (n + 1)R \). Therefore, the potential energy at this height is:
\[ U_{\text{final}} = -\frac{GMm}{(n + 1)R} \]

3. **Change in Potential Energy (ΔU)**:
The change in potential energy is given by:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we have:
\[ \Delta U = \left(-\frac{GMm}{(n + 1)R}\right) - \left(-\frac{GMm}{R}\right) \]
Simplifying this gives:
\[ \Delta U = -\frac{GMm}{(n + 1)R} + \frac{GMm}{R} = GMm \left(\frac{1}{R} - \frac{1}{(n + 1)R}\right) \]
\[ = GMm \left(\frac{(n + 1) - 1}{(n + 1)R}\right) = \frac{nGMm}{(n + 1)R} \]

4. **Expressing in terms of \( g \)**:
We know that \( g = \frac{GM}{R} \), so substituting this in gives:
\[ \Delta U = \frac{nmg}{(n + 1)} \]

Thus, the increase in potential energy is:
\[ \Delta U = \frac{n}{n + 1} mgR \]

### Part (ii): Velocity to Reach Height \( nR \)

1. **Using Conservation of Energy**:
The change in kinetic energy must equal the change in potential energy. The change in kinetic energy (ΔK) is given by:
\[ \Delta K = \frac{1}{2} mv^2 - 0 = \frac{1}{2} mv^2 \]

2. **Setting ΔK equal to ΔU**:
\[ \frac{1}{2} mv^2 = \Delta U \]
Substituting the expression for ΔU we found:
\[ \frac{1}{2} mv^2 = \frac{n}{n + 1} mgR \]

3. **Solving for v**:
Dividing both sides by \( m \) and multiplying by 2 gives:
\[ v^2 = \frac{2n}{n + 1} gR \]
Taking the square root:
\[ v = \sqrt{\frac{2ngR}{n + 1}} \]

Thus, the required velocity with which the body must be projected is:
\[ v = \sqrt{\frac{2ngR}{n + 1}} \]

### Summary of Results:
(i) The increase in potential energy is \( \frac{n}{n + 1} mgR \).

(ii) The velocity with which it must be projected is \( \sqrt{\frac{2ngR}{n + 1}} \).