Trajectory of a projectile is given by `y = alphax + betax^2`. Then find `(alpha + beta)`.

Answer

To solve the problem, we need to find the sum of the coefficients \( \alpha \) and \( \beta \) from the given projectile trajectory equation \( y = \alpha x + \beta x^2 \).

### Step-by-Step Solution:

1. **Identify the Given Equation**:
The trajectory of the projectile is given by:
\[ y = \alpha x + \beta x^2 \]

2. **Recall the General Equation of Projectile Motion**:
The general equation for the trajectory of a projectile is:
\[ y = x \tan \theta + \frac{g x^2}{2u^2 \cos^2 \theta} \]
where:
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity,
- \( u \) is the initial velocity.

3. **Compare the Two Equations**:
We will compare the coefficients of \( x \) and \( x^2 \) in both equations:
- From the term \( \alpha x \), we can equate it to \( x \tan \theta \):
\[ \alpha = \tan \theta \]
- From the term \( \beta x^2 \), we can equate it to \( \frac{g x^2}{2u^2 \cos^2 \theta} \):
\[ \beta = \frac{g}{2u^2 \cos^2 \theta} \]

4. **Find \( \alpha + \beta \)**:
Now, we need to find \( \alpha + \beta \):
\[ \alpha + \beta = \tan \theta + \frac{g}{2u^2 \cos^2 \theta} \]

5. **Final Answer**:
Thus, the value of \( \alpha + \beta \) is:
\[ \alpha + \beta = \tan \theta + \frac{g}{2u^2 \cos^2 \theta} \]