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. 0.079872685185185
B. 2.8472337962963
C. 80
D. 40

Answer

To solve the problem, we need to find the ratio \( A:B \) from the trajectory equation \( h = Ax - Bx^2 \), where \( h \) is the height, \( x \) is the horizontal distance, and \( A \) and \( B \) are constants derived from the projectile motion of an object projected at an angle.

### Step 1: Identify the parameters
Given:
- Initial velocity \( u = 20 \, \text{m/s} \)
- Angle of projection \( \theta = 45^\circ \)

### Step 2: Find \( A \)
From the general trajectory equation of projectile motion:
\[ h = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
Comparing this with the given equation \( h = Ax - Bx^2 \), we can identify:
- \( A = \tan \theta \)

Since \( \theta = 45^\circ \):
\[ A = \tan 45^\circ = 1 \]

### Step 3: Find \( B \)
Next, we need to find \( B \):
From the general trajectory equation, the coefficient of \( x^2 \) is:
\[ B = \frac{g}{2 u^2 \cos^2 \theta} \]
Substituting the known values:
- \( g = 10 \, \text{m/s}^2 \) (approximating for simplicity)
- \( u = 20 \, \text{m/s} \)
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

Now, substituting these values into the equation for \( B \):
\[ B = \frac{10}{2 \cdot (20)^2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2} \]
Calculating \( \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \):
\[ B = \frac{10}{2 \cdot 400 \cdot \frac{1}{2}} = \frac{10}{400} = \frac{1}{40} \]

### Step 4: Find the ratio \( A:B \)
Now we have:
- \( A = 1 \)
- \( B = \frac{1}{40} \)

The ratio \( A:B \) is:
\[ \frac{A}{B} = \frac{1}{\frac{1}{40}} = 40 \]

### Final Answer
Thus, the ratio \( A:B \) is \( 40:1 \).

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