From a point on the ground a particle is projected with initial velocity `u`,such that its horizontal range is maximum. The magnitude of average velocity during its ascent.

Options

A. `sqrt(5u)/(2sqrt2)`
B. `(5u)/4`
C. `sqrt3/(2sqrt2)`
D. None

Explanation

To solve the problem, we need to find the magnitude of the average velocity during the ascent of a projectile that is projected with an initial velocity \( u \) for maximum horizontal range.

### Step-by-Step Solution:

1. **Understanding Maximum Range**:
- The horizontal range \( R \) of a projectile is maximized when the angle of projection \( \theta \) is \( 45^\circ \).
- The formula for maximum range is given by:
\[ R_{\text{max}} = \frac{u^2 \sin 2\theta}{g} \]
- For \( \theta = 45^\circ \), \( \sin 90^\circ = 1 \), thus:
\[ R_{\text{max}} = \frac{u^2}{g} \]

2. **Finding the Displacement During Ascent**:
- The maximum height \( h \) reached by the projectile can be calculated using:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
- For \( \theta = 45^\circ \):
\[ h = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

3. **Calculating the Horizontal Displacement**:
- The horizontal displacement at the peak (point P) during ascent is half of the maximum range:
\[ x = \frac{R_{\text{max}}}{2} = \frac{u^2}{2g} \]

4. **Finding the Total Displacement**:
- The total displacement \( d \) from the point of projection to point P (the peak) can be calculated using the Pythagorean theorem:
\[ d = \sqrt{x^2 + h^2} \]
- Substituting the values:
\[ d = \sqrt{\left(\frac{u^2}{2g}\right)^2 + \left(\frac{u^2}{4g}\right)^2} \]
- Simplifying:
\[ d = \sqrt{\frac{u^4}{4g^2} + \frac{u^4}{16g^2}} = \sqrt{\frac{4u^4 + u^4}{16g^2}} = \sqrt{\frac{5u^4}{16g^2}} = \frac{u^2 \sqrt{5}}{4g} \]

5. **Finding the Time of Ascent**:
- The time taken to reach the maximum height can be calculated using:
\[ t = \frac{u \sin \theta}{g} = \frac{u \cdot \frac{1}{\sqrt{2}}}{g} = \frac{u}{\sqrt{2}g} \]

6. **Calculating Average Velocity**:
- The average velocity \( V_{\text{avg}} \) during ascent is given by:
\[ V_{\text{avg}} = \frac{\text{Displacement}}{\text{Time}} = \frac{d}{t} \]
- Substituting the values:
\[ V_{\text{avg}} = \frac{\frac{u^2 \sqrt{5}}{4g}}{\frac{u}{\sqrt{2}g}} = \frac{u^2 \sqrt{5}}{4g} \cdot \frac{\sqrt{2}g}{u} = \frac{u \sqrt{10}}{4} \]

### Final Answer:
Thus, the magnitude of the average velocity during ascent is:
\[ V_{\text{avg}} = \frac{u \sqrt{10}}{4} \text{ m/s} \]