In the above, the change in speed is:-

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 speed of a particle fired with an initial velocity \( u \) at an angle \( \theta \) with the horizontal, we can follow these steps:

### Step 1: Understand the Initial Velocity Components
The initial velocity \( u \) 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 Motion at the Highest Point
At the highest point of the projectile's trajectory:
- The vertical component of the velocity \( v_y \) becomes 0 (since the particle momentarily stops moving upward before descending).
- The horizontal component \( v_x \) remains unchanged throughout the motion because there is no horizontal acceleration (assuming air resistance is negligible).

Thus, at the highest point:
- \( v_x = u \cos \theta \)
- \( v_y = 0 \)

### Step 3: Calculate the Speed at the Highest Point
The speed at the highest point is given by the magnitude of the velocity vector, which is only the horizontal component at this point:
\[ v = v_x = u \cos \theta \]

### Step 4: Calculate the Change in Speed
The change in speed is defined as the final speed minus the initial speed. The initial speed is the magnitude of the initial velocity \( u \):
\[ \text{Change in speed} = v - u \]
Substituting the values we found:
\[ \text{Change in speed} = u \cos \theta - u \]

### Step 5: Simplify the Expression
We can factor out \( u \) from the expression:
\[ \text{Change in speed} = u (\cos \theta - 1) \]

### Final Answer
The change in speed of the particle when it reaches the highest point is:
\[ \text{Change in speed} = u (\cos \theta - 1) \]

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