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 is

Options

A. `(sqrt(5)u)/(2)`
B. `(sqrt(5)u)/(2sqrt(2))`
C. `(5u)/(2sqrt(2))`
D. `(sqrt(2)u)/(sqrt(5))`

Answer

To find the magnitude of the average velocity during the ascent of a particle projected with initial velocity \( u \) at an angle that maximizes its horizontal range, we can follow these steps:

### Step 1: Understand the conditions for maximum range
For a projectile launched from the ground, the angle that maximizes the horizontal range is \( \theta = 45^\circ \).

### Step 2: Calculate the time of flight
The time of flight \( T \) for a projectile launched at an angle \( \theta \) is given by:
\[ T = \frac{2u \sin \theta}{g} \]
For \( \theta = 45^\circ \), \( \sin 45^\circ = \frac{1}{\sqrt{2}} \):
\[ T = \frac{2u \cdot \frac{1}{\sqrt{2}}}{g} = \frac{u\sqrt{2}}{g} \]

### Step 3: Calculate the maximum height
The maximum height \( H \) reached by the projectile can be calculated using:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
For \( \theta = 45^\circ \):
\[ H = \frac{u^2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 4: Calculate the average velocity during ascent
The average velocity \( V_{avg} \) during the ascent is given by the formula:
\[ V_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} \]
During ascent, the total displacement is the maximum height \( H \), and the time taken to reach this height is \( \frac{T}{2} \):
\[ V_{avg} = \frac{H}{T/2} = \frac{H \cdot 2}{T} \]
Substituting the values of \( H \) and \( T \):
\[ V_{avg} = \frac{\frac{u^2}{4g} \cdot 2}{\frac{u\sqrt{2}}{g}} = \frac{u^2}{2g} \cdot \frac{g}{u\sqrt{2}} = \frac{u}{2\sqrt{2}} \]

### Step 5: Final Result
Thus, the magnitude of the average velocity during ascent is:
\[ V_{avg} = \frac{u}{2\sqrt{2}} \]