A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively

Options

A. (a)g,g
B. (b)g-a,g-a
C. (c)g-a,g
D. (d)a,g

Explanation

To solve the problem, we need to analyze the situation of a ball being dropped inside a downward-accelerating lift. We will determine the acceleration of the ball as observed by both the man in the lift and the man standing stationary on the ground.

### Step-by-step Solution:

1. **Identify the accelerations involved**:
- The lift is moving downward with an acceleration \( a \).
- The acceleration due to gravity \( g \) acts downward on the ball.

2. **Acceleration of the ball as observed by the man in the lift**:
- When the man in the lift drops the ball, he will observe the ball's acceleration relative to himself.
- Since the lift is accelerating downward with acceleration \( a \), the ball, once dropped, will also experience the same gravitational pull \( g \) downward.
- However, since the lift is accelerating downward, the effective acceleration of the ball relative to the man in the lift is:
\[ a_{\text{ball, lift}} = g - a \]
- This means that the ball appears to accelerate upward with respect to the man in the lift.

3. **Acceleration of the ball as observed by the man on the ground**:
- For the man standing on the ground, the ball is simply falling under the influence of gravity.
- Therefore, the acceleration of the ball as observed by the man on the ground is:
\[ a_{\text{ball, ground}} = g \]

4. **Final Results**:
- The acceleration of the ball as observed by the man in the lift is \( g - a \).
- The acceleration of the ball as observed by the man on the ground is \( g \).

### Conclusion:
Thus, the accelerations of the ball as observed by the man in the lift and the man on the ground are respectively \( g - a \) and \( g \).