Question

A ball is dropped on to the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 seconds, what is the average acceleration during contact? (Take $g = 10\, m \,s^{-2}$ )

• A. $700\, m \,s^{-2}$
• B. $1400\, m \,s^{-2}$
• C. $2100\, m \,s^{-2}$
• D. $2800\, m \,s^{-2}$

Answer 1

Answer: (C)

$2100\, m \,s^{-2}$

Answer 2

$\upsilon_{1} = \sqrt{2\,gh} = \sqrt{2 ? 10 ? 10} = \sqrt{200}$ $\upsilon _{2} = -\sqrt{2\,gh} = -\sqrt{2 ? 10 ? 10} = -\sqrt{50}$ So $\Delta\upsilon = \upsilon _{1} - \upsilon _{2} = \sqrt{200}+\sqrt{50} = 3\sqrt{50} = 21$ $\therefore\quad$ Acceleration $= \frac{\Delta \upsilon }{\Delta t } = \frac{21}{0.01} = 2100\,m\,s^{-2}$