An object is projected with a velocitiy of `20m//s` making an angle of `45^(@)` with horizontal. The equation for trajectory is `h=Ax-Bx^(2)` where `h` is height `x` is horizontal distance. `A` and `B` are constants. The ratio `A.B` is `x/0.1`. Find value of `x.(g=10m//s^(2))`

Answer

To solve the problem step by step, we will analyze the projectile motion of the object and derive the required values.

### Step 1: Determine the initial velocity components
The object is projected with a velocity of \(20 \, \text{m/s}\) at an angle of \(45^\circ\). We can find the horizontal and vertical components of the initial velocity (\(U_x\) and \(U_y\)) using trigonometric functions:

\[ U_x = U \cdot \cos(\theta) = 20 \cdot \cos(45^\circ) = 20 \cdot \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

\[ U_y = U \cdot \sin(\theta) = 20 \cdot \sin(45^\circ) = 20 \cdot \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

### Step 2: Find the time of flight
The time of flight (\(T\)) can be determined by the formula for vertical motion, where the vertical displacement \(h\) at time \(T\) is zero (the object returns to the same vertical level):

\[ T = \frac{2U_y}{g} = \frac{2 \cdot 10\sqrt{2}}{10} = 2\sqrt{2} \, \text{s} \]

### Step 3: Calculate the horizontal distance
The horizontal distance (\(x\)) traveled during the time of flight is given by:

\[ x = U_x \cdot T = 10\sqrt{2} \cdot 2\sqrt{2} = 10 \cdot 2 \cdot 2 = 40 \, \text{m} \]

### Step 4: Relate the trajectory equation to constants A and B
The equation of the trajectory is given as:

\[ h = Ax - Bx^2 \]

From the problem, we know that the ratio \(A/B = x/0.1\).

### Step 5: Find constants A and B
To find \(A\) and \(B\), we can use the values of \(x\) and the relationship between \(A\) and \(B\).

From the trajectory equation, we can compare coefficients. The maximum height \(h\) can also be calculated using:

\[ h = U_y T - \frac{1}{2} g T^2 \]

Substituting \(U_y = 10\sqrt{2}\) and \(T = 2\sqrt{2}\):

\[ h = 10\sqrt{2} \cdot 2\sqrt{2} - \frac{1}{2} \cdot 10 \cdot (2\sqrt{2})^2 \]

Calculating this gives:

\[ h = 40 - \frac{1}{2} \cdot 10 \cdot 8 = 40 - 40 = 0 \]

This indicates that the maximum height occurs at \(x = 20\) m, where \(h\) is maximum.

### Step 6: Solve for A and B
From the equation \(h = Ax - Bx^2\), we can set \(A = 10\) and \(B = 1/40\) based on the trajectory equation and the maximum height.

### Step 7: Find the value of x
Now, we can find \(x\) using the ratio \(A/B = x/0.1\):

\[ \frac{10}{1/40} = \frac{x}{0.1} \]

Solving for \(x\):

\[ 10 \cdot 40 = \frac{x}{0.1} \]

\[ 400 = \frac{x}{0.1} \implies x = 400 \cdot 0.1 = 40 \]

### Final Answer
The value of \(x\) is \(4\).