At the top of the trajectory of projectile the

Options

A. Acceleration is minimum
B. Velocity is zero
C. Acceleration is maximum
D. Acceleration is g

Answer

To solve the question "At the top of the trajectory of a projectile, what is true?", we will analyze the motion of a projectile and its characteristics at the peak of its trajectory.

### Step-by-Step Solution:

1. **Understanding Projectile Motion**:
- A projectile is an object that is thrown into the air with an initial velocity at an angle to the horizontal. The motion can be analyzed in two dimensions: horizontal and vertical.

2. **Components of Velocity**:
- The initial velocity \( u \) can be broken down into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

3. **At the Top of the Trajectory**:
- At the highest point of the projectile's path, the vertical component of the velocity becomes zero. This is because the projectile momentarily stops rising before it starts to fall back down.
- Therefore, the vertical velocity \( v_y = 0 \) at the top.

4. **Horizontal Velocity**:
- The horizontal component of the velocity remains unchanged throughout the motion (assuming no air resistance). Thus, at the top, the horizontal velocity \( v_x = u \cos \theta \) is still present.

5. **Acceleration**:
- The only force acting on the projectile is gravity, which acts downwards with a constant acceleration \( g \). This acceleration does not change with height (as long as the height is much smaller than the radius of the Earth).
- Therefore, the acceleration at the top of the trajectory is equal to \( g \).

6. **Conclusion**:
- At the top of the trajectory of a projectile:
- The vertical velocity is zero.
- The horizontal velocity is \( u \cos \theta \).
- The acceleration is \( g \), which is constant and directed downwards.

### Answer:
The correct statement is: **Acceleration is \( g \)**.