What is the change in velocity in the above question?

Explanation

To find the change in velocity of a projectile that is launched with an initial velocity \( u \) at an angle \( \alpha \) with the horizontal, we can follow these steps:

### Step 1: Identify Initial and Final Velocities
- The initial velocity \( \mathbf{u} \) can be broken down into its components:
- Horizontal component: \( u_x = u \cos \alpha \)
- Vertical component: \( u_y = u \sin \alpha \)

- At the point of impact (point B), the horizontal component of the velocity remains the same since there is no horizontal acceleration:
- Final horizontal component: \( v_x = u \cos \alpha \)

### Step 2: Calculate Final Vertical Velocity
- The vertical component of the velocity at point B can be calculated using the first equation of motion:
\[ v_y = u_y + a_y \cdot t \]
where \( a_y = -g \) (acceleration due to gravity) and \( t \) is the time of flight.

- The time of flight \( t \) for a projectile is given by:
\[ t = \frac{2u \sin \alpha}{g} \]

- Substituting \( u_y \) and \( t \) into the equation for \( v_y \):
\[ v_y = u \sin \alpha - g \cdot \frac{2u \sin \alpha}{g} \]
Simplifying this gives:
\[ v_y = u \sin \alpha - 2u \sin \alpha = -u \sin \alpha \]

### Step 3: Determine Change in Velocity
- The change in velocity \( \Delta \mathbf{v} \) is given by:
\[ \Delta \mathbf{v} = \mathbf{v} - \mathbf{u} \]
This can be expressed in terms of its components:
\[ \Delta \mathbf{v} = (v_x - u_x) \hat{i} + (v_y - u_y) \hat{j} \]

- Substituting the values:
- \( v_x = u \cos \alpha \) and \( u_x = u \cos \alpha \) gives:
\[ v_x - u_x = 0 \]

- For the vertical component:
\[ v_y - u_y = (-u \sin \alpha) - (u \sin \alpha) = -2u \sin \alpha \]

- Therefore, the change in velocity is:
\[ \Delta \mathbf{v} = 0 \hat{i} - 2u \sin \alpha \hat{j} \]
or simply:
\[ \Delta \mathbf{v} = -2u \sin \alpha \hat{j} \]

### Final Result
The change in velocity of the projectile is:
\[ \Delta \mathbf{v} = -2u \sin \alpha \hat{j} \]

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