A tennis ball is dropped on to the floor from a height of \(9.8\,m\). It rebounds to a height \(5.0\,m\). Ball comes in contact with the floor for \(0.2\,s\). The average acceleration during contact is \(..........ms ^{-2}\). [Given \(g =10\,ms ^{-2}\) ]

Explanation:
1. Find the velocity of the ball just before it strikes the floor (\( v_i \)):
Using the third equation of motion for a freely falling body:
\( v_i = \sqrt{2gh_i} \)
Given initial height, \( h_i = 9.8 \text{ m} \) and \( g = 10 \text{ ms}^{-2} \):
\( v_i = \sqrt{2 \times 10 \times 9.8} = \sqrt{196} = 14 \text{ m/s} \) (directed downwards)

2. Find the velocity of the ball just after it rebounds from the floor (\( v_f \)):
Using the same principle for the rebound height:
\( v_f = \sqrt{2gh_f} \)
Given final height, \( h_f = 5.0 \text{ m} \):
\( v_f = \sqrt{2 \times 10 \times 5.0} = \sqrt{100} = 10 \text{ m/s} \) (directed upwards)

3. Calculate the average acceleration (\( a_{avg} \)):
Let us choose the upward direction as positive and the downward direction as negative:
Initial velocity, \( v_i = -14 \text{ m/s} \)
Final velocity, \( v_f = +10 \text{ m/s} \)

The change in velocity (\( \Delta v \)) is given by:
\( \Delta v = v_f - v_i = 10 - (-14) = 24 \text{ m/s} \)

The average acceleration during the contact time interval \( \Delta t = 0.2 \text{ s} \) is:
\( a_{avg} = \frac{\Delta v}{\Delta t} = \frac{24}{0.2} = 120 \text{ m/s}^2 \)

Hence, Option A is the correct choice.