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 regarding the projectile motion at the top of its trajectory, we will analyze the components of velocity and acceleration at that point.

### Step-by-Step Solution:

1. **Understanding the Projectile Motion**:
- A projectile is launched with an initial velocity \( u \) at an angle \( \theta \) with the horizontal. The motion can be analyzed in two dimensions: horizontal (x-direction) and vertical (y-direction).

2. **Components of Initial Velocity**:
- The initial velocity 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 top of the trajectory), the vertical component of the velocity becomes zero. This is because the projectile has reached its maximum height and is about to start descending.

4. **Velocity at the Highest Point**:
- While the vertical component of the velocity is zero at the highest point, the horizontal component remains unchanged throughout the motion (since there is no horizontal acceleration). Thus, the net velocity at the top is:
\[ v = u \cos \theta \]
- Therefore, the velocity is not zero; it has a horizontal component.

5. **Acceleration at the Highest Point**:
- The only force acting on the projectile is gravity, which acts downward with an acceleration of \( g \). This acceleration remains constant throughout the motion, including at the highest point.
- Hence, the acceleration at the top of the trajectory is:
\[ a = -g \]
- This means the acceleration is equal to \( g \) (in magnitude) and directed downwards.

6. **Conclusion**:
- At the top of the trajectory:
- The vertical component of velocity is zero.
- The horizontal component of velocity is \( u \cos \theta \).
- The acceleration is constant and equal to \( g \) (downward).

### Final Answer:
- The acceleration at the top of the trajectory is \( g \) (downward), and the velocity is not zero; it is \( u \cos \theta \) (horizontal).