Frame of the man standing on the ground

The ground frame is an inertial frame.

After the ball is dropped, the only force acting on it is gravity. Hence its acceleration with respect to the ground is $a_{\text{ball, ground}} = g \quad \text{downward}.$

Frame of the man inside the lift

The lift is moving downward with acceleration $a$, so the lift frame is a non-inertial frame.

In the lift frame, we introduce a pseudo force opposite to the lift's acceleration, i.e. upward.

• Real force on ball: weight $mg$ downward
• Pseudo force in lift frame: $ma$ upward

Therefore net force in the lift frame is $F_{\text{net, lift}} = mg - ma = m(g-a) \quad \text{downward}.$

So the acceleration of the ball as observed by the man in the lift is $a_{\text{ball, lift}} = g-a \quad \text{downward}.$

Matching with options

Respectively, the accelerations observed by:

• man in the lift: $g-a$
• man on the ground: $g$

Hence the correct option is $\boxed{\text{C: } (g-a,\ g)}.$

Comparison with stored correct answer

Stored correct answer is C, which matches our derived answer.