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 in a lift that is moving downward with an acceleration \( a \). We will find the acceleration of the ball as observed by two different observers: the man inside the lift and the man standing stationary on the ground.

### Step-by-Step Solution:

1. **Understanding the Frame of Reference**:
- The lift is moving downward with an acceleration \( a \).
- The man inside the lift drops the ball. We need to determine the acceleration of the ball from two perspectives: the man in the lift and the man on the ground.

2. **Acceleration of the Ball as Observed by the Man in the Lift**:
- When the man inside the lift drops the ball, he is in a non-inertial frame of reference (the lift is accelerating downwards).
- To the man in the lift, the ball appears to be accelerating upwards because the lift is accelerating downwards.
- The effective acceleration of the ball relative to the man in the lift is:
\[ a_{\text{ball (lift)}} = g - a \]
- Here, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). The ball experiences a downward acceleration of \( g \), but since the lift is accelerating downwards with \( a \), the relative acceleration observed by the man in the lift is \( g - a \).

3. **Acceleration of the Ball as Observed by the Man on the Ground**:
- The man on the ground is in an inertial frame of reference. He observes the ball falling freely under the influence of gravity.
- Therefore, the acceleration of the ball as observed by the man on the ground is simply:
\[ a_{\text{ball (ground)}} = g \]

4. **Final Answers**:
- 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 \( g \).

### Summary of Results:
- Acceleration of the ball as observed by the man in the lift: \( g - a \)
- Acceleration of the ball as observed by the man on the ground: \( g \)