Two balls A and B are projected from the same position with horizontal velocity `20 ms^(-1)` and `5 ms^(-1)` as shown. The separation between these two bodies at the moment their velocities become perpendicular to each other is x m. The value of x is

Answer

To solve the problem, we need to determine the separation between two balls A and B at the moment their velocities become perpendicular to each other. Here’s a step-by-step solution:

### Step 1: Define the velocities of the balls
- Ball A is projected with a horizontal velocity \( v_A = 20 \, \text{m/s} \).
- Ball B is projected with a horizontal velocity \( v_B = 5 \, \text{m/s} \).

### Step 2: Write the velocity vectors of both balls
Assuming both balls are projected in the same horizontal direction:
- The velocity vector of ball A can be expressed as:
\[ \vec{v_A} = 20 \hat{i} - 5t \hat{j} \]
- The velocity vector of ball B can be expressed as:
\[ \vec{v_B} = 5 \hat{i} - 5t \hat{j} \]

### Step 3: Determine the condition for perpendicular velocities
The velocities are perpendicular when their dot product is zero:
\[ \vec{v_A} \cdot \vec{v_B} = 0 \]
Calculating the dot product:
\[ (20 \hat{i} - 5t \hat{j}) \cdot (5 \hat{i} - 5t \hat{j}) = 20 \times 5 + (-5t)(-5t) = 100 + 25t^2 = 0 \]
This simplifies to:
\[ 100 + 25t^2 = 0 \]
Since \(t^2\) cannot be negative, we need to consider the case where the balls are projected in opposite directions.

### Step 4: Adjust the velocity vectors for opposite directions
If we assume ball B is projected in the opposite direction:
- The velocity vector of ball A remains the same:
\[ \vec{v_A} = 20 \hat{i} - 5t \hat{j} \]
- The velocity vector of ball B becomes:
\[ \vec{v_B} = -5 \hat{i} - 5t \hat{j} \]

### Step 5: Calculate the new dot product
Now, we calculate the dot product again:
\[ \vec{v_A} \cdot \vec{v_B} = (20 \hat{i} - 5t \hat{j}) \cdot (-5 \hat{i} - 5t \hat{j}) = 20 \times (-5) + (-5t)(-5t) = -100 + 25t^2 \]
Setting this to zero for perpendicularity:
\[ -100 + 25t^2 = 0 \]
Solving for \(t^2\):
\[ 25t^2 = 100 \implies t^2 = 4 \implies t = 2 \, \text{s} \]

### Step 6: Calculate the position vectors at \(t = 2\)
Now we find the position of each ball at \(t = 2\):
- For ball A:
\[ \vec{r_A} = \vec{v_A} \cdot t = (20 \hat{i} - 5(2) \hat{j}) \cdot 2 = (20 \cdot 2) \hat{i} - (10) \hat{j} = 40 \hat{i} - 10 \hat{j} \]
- For ball B:
\[ \vec{r_B} = \vec{v_B} \cdot t = (-5 \hat{i} - 5(2) \hat{j}) \cdot 2 = (-5 \cdot 2) \hat{i} - (10) \hat{j} = -10 \hat{i} - 10 \hat{j} \]

### Step 7: Find the separation between the two balls
The separation \(x\) is given by the magnitude of the difference of their position vectors:
\[ \vec{r_A} - \vec{r_B} = (40 \hat{i} - 10 \hat{j}) - (-10 \hat{i} - 10 \hat{j}) = (40 + 10) \hat{i} + (0) \hat{j} = 50 \hat{i} \]
The magnitude of this vector is:
\[ x = |50 \hat{i}| = 50 \, \text{m} \]

### Final Answer
The separation between the two bodies at the moment their velocities become perpendicular is:
\[ \boxed{50 \, \text{m}} \]