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. `u`
C. `u sin theta`
D. `(u cos theta-u)`

Explanation

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

### Step 1: Break down the initial velocity into components
The initial velocity \(u\) can be resolved into horizontal and vertical components:
- Horizontal component: \(u_x = u \cos \theta\)
- Vertical component: \(u_y = u \sin \theta\)

### Step 2: Analyze the motion at the highest point
At the highest point of the projectile's trajectory:
- The vertical component of the velocity becomes zero (\(v_y = 0\)) because the particle momentarily stops moving upwards before descending.
- The horizontal component of the velocity remains constant throughout the motion, so \(v_x = u \cos \theta\).

### 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:
\[ \vec{v} = u \cos \theta \hat{i} + 0 \hat{j} = u \cos \theta \hat{i} \]

### 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 \(\vec{u}\):
\[ \Delta \vec{v} = (u \cos \theta \hat{i}) - (u \cos \theta \hat{i} + u \sin \theta \hat{j}) \]
This simplifies to:
\[ \Delta \vec{v} = -u \sin \theta \hat{j} \]

### Step 5: Determine the magnitude and direction of the change in velocity
The magnitude of the change in velocity is:
\[ |\Delta \vec{v}| = u \sin \theta \]
The direction is downward along the negative \(y\)-axis, which can be represented as \(-\hat{j}\).

### Final Answer
Thus, the change in velocity when the particle is at the highest point is:
\[ \Delta \vec{v} = -u \sin \theta \hat{j} \]
with a magnitude of \(u \sin \theta\) and direction \(-\hat{j}\).

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