A particle is projected up the inclined such that its component of velocity along the incline is `10 m//s`. Time of flight is `2` sec and maximum height above the incline is `5m`. Then velocity of projection will be

Options

A. `10 m//s`
B. `10 sqrt(2) m//s`
C. `5sqrt(5) m//s`
D. none

Explanation

To solve the problem step by step, we will analyze the motion of the particle projected up an inclined plane.

### Step 1: Identify the Given Information
- Component of velocity along the incline, \( V_{\parallel} = 10 \, \text{m/s} \)
- Time of flight, \( t = 2 \, \text{s} \)
- Maximum height above the incline, \( h = 5 \, \text{m} \)

### Step 2: Analyze the Motion
The motion can be divided into two components:
1. Along the incline (parallel to the incline)
2. Perpendicular to the incline (vertical direction)

### Step 3: Time to Reach Maximum Height
Since the total time of flight is 2 seconds, the time taken to reach the maximum height is half of that:
\[ t_{\text{up}} = \frac{t}{2} = 1 \, \text{s} \]

### Step 4: Use the Vertical Motion Equation
At maximum height, the vertical component of the velocity becomes zero. The vertical component of the initial velocity can be expressed as:
\[ V_{\text{vertical}} = V \sin \theta \]
Using the equation of motion:
\[ V = u + at \]
Where:
- \( V = 0 \) (at maximum height)
- \( a = -g \) (acceleration due to gravity)
- \( t = 1 \, \text{s} \)

Thus, we have:
\[ 0 = V \sin \theta - g \cdot t_{\text{up}} \]
\[ V \sin \theta = g \cdot 1 \]
\[ V \sin \theta = g \]

### Step 5: Use the Maximum Height Equation
The maximum height \( h \) can also be expressed using the vertical component of the initial velocity:
\[ h = V \sin \theta \cdot t_{\text{up}} - \frac{1}{2} g t_{\text{up}}^2 \]
Substituting \( t_{\text{up}} = 1 \, \text{s} \):
\[ 5 = V \sin \theta \cdot 1 - \frac{1}{2} g \cdot 1^2 \]
\[ 5 = V \sin \theta - \frac{g}{2} \]
Substituting \( V \sin \theta = g \):
\[ 5 = g - \frac{g}{2} \]
\[ 5 = \frac{g}{2} \]
\[ g = 10 \, \text{m/s}^2 \]

### Step 6: Use the Horizontal Component of Velocity
We know that:
\[ V \cos \theta = 10 \, \text{m/s} \]

### Step 7: Combine the Equations
Now we have two equations:
1. \( V \sin \theta = g = 10 \, \text{m/s}^2 \)
2. \( V \cos \theta = 10 \, \text{m/s} \)

### Step 8: Square and Add the Equations
Squaring both equations and adding them:
\[ (V \sin \theta)^2 + (V \cos \theta)^2 = (10)^2 + (10)^2 \]
\[ V^2 (\sin^2 \theta + \cos^2 \theta) = 100 + 100 \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ V^2 = 200 \]
\[ V = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \]

### Final Answer
The velocity of projection \( V \) is:
\[ \boxed{10\sqrt{2} \, \text{m/s}} \]