A projectill is projected at an angle `(alpha gt 45^(@))` with an initial velocity u. The time t, at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is :-

Options

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

Explanation

To solve the problem of finding the time \( t \) at which the magnitude of horizontal velocity equals the magnitude of 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 Components of Initial Velocity**:
The initial velocity \( u \) can be broken down into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \alpha \)
- Vertical component: \( u_y = u \sin \alpha \)

2. **Determine the Horizontal Velocity**:
In projectile motion, the horizontal velocity remains constant throughout the motion because there is no horizontal acceleration. Thus:
\[ v_x = u \cos \alpha \]

3. **Determine the Vertical Velocity**:
The vertical velocity changes due to gravitational acceleration. The vertical velocity at any time \( t \) can be expressed as:
\[ v_y = u_y - g t = u \sin \alpha - g t \]

4. **Set Horizontal Velocity Equal to Vertical Velocity**:
We need to find the time \( t \) when the magnitudes of the horizontal and vertical velocities are equal:
\[ u \cos \alpha = u \sin \alpha - g t \]

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

6. **Factor Out \( u \)**:
Factoring out \( u \) from the right side:
\[ g t = u (\sin \alpha - \cos \alpha) \]

7. **Solve for Time \( t \)**:
Finally, we can solve for \( t \):
\[ t = \frac{u (\sin \alpha - \cos \alpha)}{g} \]

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