Answer:
[tex]\large \boxed{\text{98 m$\cdot$s}^{-2}}[/tex]
Explanation:
The formula for the velocity of the ball is
[tex]v = \sqrt{2gh}[/tex]
1. Velocity at time of impact
[tex]v = -\sqrt{2 \times 9.807 \times 1.20} = -\sqrt{23.54} = -\textbf{4.85 m/s}[/tex]
2. Velocity on rebound
The ball has enough upward velocity to reach a height of 0.86 m.
[tex]v = \sqrt{2 \times 9.807 \times 0.86} = \sqrt{16.87} =\textbf{4.11 m/s}[/tex]
3. Acceleration
[tex]a = \dfrac{\Delta v}{\Delta t} = \dfrac{4.11 - (-4.85)}{ 0.091} = \dfrac{8.96 }{0.091} =\textbf{98 m$\cdot$s}^{\mathbf{-2}}\\\\\text{The acceleration while the ball is in contact with the floor is $\large \boxed{\textbf{98 m$\cdot$s}^{\mathbf{-2}}}$}[/tex]
