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 solve the problem of finding the time \( t \) at which the horizontal velocity equals the vertical velocity for a projectile launched at an angle \( \alpha \) with an initial velocity \( u \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Horizontal and Vertical Components of Velocity:**
- The horizontal component of the initial velocity \( v_{x} \) is given by:
\[ v_{x} = u \cos \alpha \]
- The vertical component of the initial velocity \( v_{y} \) is given by:
\[ v_{y} = u \sin \alpha \]

2. **Understand the Behavior of Velocities Over Time:**
- The horizontal velocity remains constant throughout the motion since there is no horizontal acceleration:
\[ v_{x} = u \cos \alpha \]
- The vertical velocity changes over time due to gravitational acceleration \( g \):
\[ v_{y}(t) = u \sin \alpha - g t \]

3. **Set the Horizontal and Vertical Velocities Equal:**
- We need to find the time \( t \) when the horizontal velocity equals the vertical velocity:
\[ u \cos \alpha = u \sin \alpha - g t \]

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

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

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