A projectile is projected at an angle `alpha` with an initial velocity u. The time t, at which its horizontal velocity will equal the vertical velocity for the first time

Options

A. `t = u ( cos alpha - sin alpha ) //g `
B. `t =u ( cos alpha + sin alpha ) //g `
C. `t =u ( sin alpha = cos alpha ) //g `
D. `t=u ( sin ^(2) alpha - cos ^(2) alpha ) //g`

Answer

To find the time \( t \) at which the horizontal velocity equals the vertical velocity for a projectile projected at an angle \( \alpha \) with an initial velocity \( u \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Velocity**:
The initial velocity \( u \) can be resolved into horizontal and vertical components:
- Horizontal velocity \( v_x = u \cos \alpha \) (constant)
- Vertical velocity \( v_y = u \sin \alpha - g t \) (changes with time due to gravity)

2. **Set Horizontal Velocity Equal to Vertical Velocity**:
We need to find the time \( t \) when the horizontal velocity equals the vertical velocity:
\[ v_x = v_y \]
Substituting the expressions for \( v_x \) and \( v_y \):
\[ u \cos \alpha = u \sin \alpha - g t \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ g t = u \sin \alpha - u \cos \alpha \]

4. **Solve for Time \( t \)**:
Now, we can isolate \( t \):
\[ t = \frac{u \sin \alpha - u \cos \alpha}{g} \]
This can be simplified to:
\[ t = \frac{u (\sin \alpha - \cos \alpha)}{g} \]

5. **Final Expression**:
Thus, the time \( t \) at which the horizontal velocity equals the vertical velocity for the first time is:
\[ t = \frac{u (\sin \alpha - \cos \alpha)}{g} \]

### Final Answer:
The time \( t \) at which the horizontal velocity equals the vertical velocity is:
\[ t = \frac{u (\sin \alpha - \cos \alpha)}{g} \]