Projectile motion is a combination of two one-dimensional motion: one in horizontal and other in vertical direction. Motion in 2D means in a plane. Necessary condition for 2D motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be `0^@ltthetalt180^@`. During the projectile motion, the horizontal component of velocity ramains unchanged but the vertical component of velocity is time dependent. Now answer the following questions:
A particle is projected from the origin in the x-y plane. The acceleration of particle in negative y-direction is `alpha`. If equation of path of the particle is `y = ax - bx^2`, then initial velocity of the particle is

Options

A. `sqrt(alpha/(2b))`
B. `sqrt((alpha(1+a^2))/(2b))`
C. `sqrt(alpha/(a^2))`
D. `sqrt((alphab)/(a^2))`

Explanation

To find the initial velocity of a particle projected from the origin in the x-y plane, given the equation of its path as \( y = ax - bx^2 \) and the acceleration in the negative y-direction as \( -\alpha \), we can follow these steps:

### Step 1: Differentiate the Equation of the Path
The equation of the path is given by:
\[ y = ax - bx^2 \]
To find the vertical component of the velocity, we differentiate \( y \) with respect to time \( t \):
\[ \frac{dy}{dt} = a \frac{dx}{dt} - 2b x \frac{dx}{dt} \]
Let \( V_x = \frac{dx}{dt} \) (horizontal component of velocity) and \( V_y = \frac{dy}{dt} \) (vertical component of velocity):
\[ V_y = a V_x - 2b x V_x \]

### Step 2: Evaluate at the Initial Point
At the initial point, \( x = 0 \):
\[ V_y = a V_x \]
This means that at the origin, the vertical component of the velocity is directly proportional to the horizontal component.

### Step 3: Differentiate Again to Find Acceleration
Next, we differentiate \( V_y \) with respect to time to find the vertical acceleration:
\[ \frac{dV_y}{dt} = \frac{d}{dt}(a V_x - 2b x V_x) \]
Using the product rule:
\[ \frac{dV_y}{dt} = a \frac{dV_x}{dt} - 2b \left( V_x \frac{dV_x}{dt} + x \frac{dV_x}{dt} \right) \]
At \( x = 0 \), the term involving \( x \) vanishes:
\[ \frac{dV_y}{dt} = a \frac{dV_x}{dt} - 2b V_x \frac{dV_x}{dt} \]

### Step 4: Relate to Given Acceleration
The vertical acceleration is given as \( -\alpha \):
\[ -\alpha = a \frac{dV_x}{dt} - 2b V_x \frac{dV_x}{dt} \]
This can be rearranged to:
\[ -\alpha = (a - 2b V_x) \frac{dV_x}{dt} \]

### Step 5: Solve for \( V_x \)
Since the horizontal acceleration is zero, we can assume \( \frac{dV_x}{dt} = 0 \). Thus, we can find \( V_x \) using the relationship from the vertical acceleration:
\[ -\alpha = -2b V_x^2 \Rightarrow \alpha = 2b V_x^2 \]
From this, we can express \( V_x \):
\[ V_x^2 = \frac{\alpha}{2b} \]

### Step 6: Find the Initial Velocity
The initial velocity \( V \) can be expressed as:
\[ V = \sqrt{V_x^2 + V_y^2} \]
Substituting \( V_y = a V_x \):
\[ V = \sqrt{V_x^2 + (a V_x)^2} = \sqrt{V_x^2(1 + a^2)} \]
Thus, substituting \( V_x^2 \):
\[ V = V_x \sqrt{1 + a^2} = \sqrt{\frac{\alpha}{2b}} \sqrt{1 + a^2} \]
Finally, we have:
\[ V = \sqrt{\frac{\alpha}{2b}(1 + a^2)} \]

### Final Answer
The initial velocity of the particle is:
\[ V = \sqrt{\frac{\alpha}{2b}(1 + a^2)} \]