A body is projected with a speed (u) at an angle to the horizontal to have maximum range. What is its velocity at the highest point ?

Options

A. zero
B. `u sqrt(2)`
C. `u`
D. `(u)/(sqrt(2))`

Explanation

To find the velocity of a body projected at an angle to the horizontal at its highest point, we can follow these steps:

### Step 1: Understand the Components of Velocity
When a body is projected at an angle, its initial velocity (u) can be broken down into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

For maximum range, the angle of projection \( \theta \) is 45 degrees.

### Step 2: Calculate the Components at 45 Degrees
At \( \theta = 45^\circ \):
- \( u_x = u \cos 45^\circ = u \cdot \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}} \)
- \( u_y = u \sin 45^\circ = u \cdot \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}} \)

### Step 3: Analyze the Motion at the Highest Point
At the highest point of its trajectory:
- The vertical component of the velocity becomes zero because the body momentarily stops rising before it starts to fall back down.
- The horizontal component remains unchanged throughout the motion (assuming no air resistance).

### Step 4: Determine the Velocity at the Highest Point
Since the vertical component of the velocity is zero at the highest point, the total velocity at this point is equal to the horizontal component:
- \( v = u_x = \frac{u}{\sqrt{2}} \)

### Conclusion
Thus, the velocity of the body at the highest point is \( \frac{u}{\sqrt{2}} \).