An object is projected with a velocity of `20(m)/(s)` making an angle of `45^@` with horizontal. The equation for the trajectory is `h=Ax-Bx^2` where h is height, x is horizontal distance, A and B are constants. The ratio A:B is (g`=ms^-2)`

Options

A. `1:5`
B. `5:1`
C. `1:40`
D. `40:1`

Explanation

To solve the problem, we need to find the ratio \( A:B \) in the trajectory equation given by \( h = Ax - Bx^2 \).

### Step-by-Step Solution:

1. **Understanding the Trajectory Equation**:
The general equation for the trajectory of a projectile is given by:
\[ y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta} x^2 \]
Here, \( y \) is the height (h), \( x \) is the horizontal distance, \( \theta \) is the angle of projection, \( g \) is the acceleration due to gravity, and \( u \) is the initial velocity.

2. **Identifying Constants**:
By comparing the general trajectory equation with the given equation \( h = Ax - Bx^2 \), we can identify:
- \( A = \tan \theta \)
- \( B = \frac{g}{2u^2 \cos^2 \theta} \)

3. **Substituting Values**:
We know:
- The initial velocity \( u = 20 \, \text{m/s} \)
- The angle \( \theta = 45^\circ \)

Now, we can calculate \( A \) and \( B \):
- For \( A \):
\[ A = \tan(45^\circ) = 1 \]
- For \( B \):
\[ B = \frac{g}{2u^2 \cos^2(45^\circ)} \]
Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \cos^2(45^\circ) = \frac{1}{2} \]
Thus,
\[ B = \frac{g}{2 \cdot (20)^2 \cdot \frac{1}{2}} = \frac{g}{400} \]

4. **Finding the Ratio \( A:B \)**:
Now, we can find the ratio \( \frac{A}{B} \):
\[ \frac{A}{B} = \frac{1}{\frac{g}{400}} = \frac{400}{g} \]

5. **Calculating the Final Ratio**:
Substituting \( g \approx 10 \, \text{m/s}^2 \) (for simplicity):
\[ \frac{A}{B} = \frac{400}{10} = 40 \]
Therefore, the ratio \( A:B \) is:
\[ A:B = 40:1 \]

### Final Answer:
The ratio \( A:B \) is \( 40:1 \).