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 ______ m s$$^{-2}$$. [Given $$g = 10$$ m s$$^{-2}$$]

When the ball is dropped from a height of $$9.8$$ m, its velocity just before hitting the floor is found using $$v^2 = 2gh$$: $$v = \sqrt{2 \times 10 \times 9.8} = \sqrt{196} = 14$$ m/s (downward).

After rebounding to a height of $$5.0$$ m, the velocity just after leaving the floor is $$v' = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10$$ m/s (upward).

Taking upward as positive, the velocity changes from $$-14$$ m/s (downward, just before contact) to $$+10$$ m/s (upward, just after contact). The average acceleration during the $$0.2$$ s contact time is $$a = \dfrac{v' - v}{\Delta t} = \dfrac{10 - (-14)}{0.2} = \dfrac{24}{0.2} = \boxed{120}$$ m s$$^{-2}$$.