A body is projected horizontally with a speed `v_(0)` find the velocity of the body when it covers equal distance in horizontal and vertical directions.

Answer

To solve the problem of finding the velocity of a body projected horizontally with an initial speed \( v_0 \) when it covers equal distances in the horizontal and vertical directions, we can follow these steps:

### Step 1: Identify the initial conditions
The body is projected horizontally, which means:
- The initial horizontal velocity \( u_x = v_0 \)
- The initial vertical velocity \( u_y = 0 \)

### Step 2: Write the equations for horizontal and vertical distances
1. **Horizontal distance \( S_x \)**:
\[ S_x = u_x \cdot t + \frac{1}{2} a_x \cdot t^2 \]
Since there is no horizontal acceleration (\( a_x = 0 \)):
\[ S_x = v_0 \cdot t \]

2. **Vertical distance \( S_y \)**:
\[ S_y = u_y \cdot t + \frac{1}{2} a_y \cdot t^2 \]
Here, \( u_y = 0 \) and the vertical acceleration \( a_y = g \) (acceleration due to gravity):
\[ S_y = 0 + \frac{1}{2} g t^2 = \frac{1}{2} g t^2 \]

### Step 3: Set the horizontal distance equal to the vertical distance
To find the time \( t \) when \( S_x = S_y \):
\[ v_0 \cdot t = \frac{1}{2} g t^2 \]

### Step 4: Solve for time \( t \)
Rearranging the equation gives:
\[ 2 v_0 t = g t^2 \]
Dividing both sides by \( t \) (assuming \( t \neq 0 \)):
\[ 2 v_0 = g t \implies t = \frac{2 v_0}{g} \]

### Step 5: Find the horizontal and vertical components of velocity at time \( t \)
1. **Horizontal component \( v_x \)**:
Since there is no horizontal acceleration:
\[ v_x = u_x = v_0 \]

2. **Vertical component \( v_y \)**:
Using the equation \( v_y = u_y + a_y t \):
\[ v_y = 0 + g \cdot t = g \cdot \left(\frac{2 v_0}{g}\right) = 2 v_0 \]

### Step 6: Calculate the resultant velocity \( v \)
The resultant velocity \( v \) can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(v_0)^2 + (2 v_0)^2} = \sqrt{v_0^2 + 4 v_0^2} = \sqrt{5 v_0^2} = v_0 \sqrt{5} \]

### Final Answer
The velocity of the body when it covers equal distances in horizontal and vertical directions is:
\[ v = v_0 \sqrt{5} \]

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