A particle is fired with velocity `u` making angle `theta` with the horizontal.What is the change in velocity when it is at the highest point?

Options

A. `u cos theta`
B. uniform retardation
C. `u sin theta`
D. `(u cos theta-u)`

Explanation

To find the change in velocity of a particle fired with an initial velocity \( u \) at an angle \( \theta \) with the horizontal when it reaches the highest point, we can follow these steps:

### Step 1: Determine the initial velocity components
The initial velocity \( \vec{u} \) can be broken down into its horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Analyze the velocity at the highest point
At the highest point of the projectile's trajectory, the vertical component of the velocity becomes zero because the particle momentarily stops moving upward before descending. Therefore, the velocity at the highest point \( \vec{v} \) is:
- Horizontal component: \( v_x = u \cos \theta \)
- Vertical component: \( v_y = 0 \)

### Step 3: Write the initial and final velocity vectors
The initial velocity vector \( \vec{u} \) can be expressed as:
\[ \vec{u} = u \cos \theta \hat{i} + u \sin \theta \hat{j} \]
The final velocity vector \( \vec{v} \) at the highest point is:
\[ \vec{v} = u \cos \theta \hat{i} + 0 \hat{j} \]

### Step 4: Calculate the change in velocity
The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v} - \vec{u} \]
Substituting the expressions for \( \vec{v} \) and \( \vec{u} \):
\[ \Delta \vec{v} = (u \cos \theta \hat{i} + 0 \hat{j}) - (u \cos \theta \hat{i} + u \sin \theta \hat{j}) \]
This simplifies to:
\[ \Delta \vec{v} = -u \sin \theta \hat{j} \]

### Step 5: Find the magnitude of the change in velocity
The magnitude of the change in velocity is:
\[ |\Delta \vec{v}| = |-u \sin \theta| = u \sin \theta \]

### Conclusion
Thus, the change in velocity when the particle is at the highest point is \( u \sin \theta \).

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