Question

A ball is dropped on 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 sec, the average acceleration during contact is

• A. 2100 $m / \sec ^ { 2 }$ downwards
• B. 2100 $m / \sec ^ { 2 }$ upwards
• C. 1400 $m / \sec ^ { 2 }$
• D. 700 $m / \sec ^ { 2 }$

Answer 1

Answer: (B)

2100 $m / \sec ^ { 2 }$ upwards

Answer 2

Velocity at the time of striking the floor,

$u = \sqrt { 2 g h _ { 1 } } = \sqrt { 2 \times 9.8 \times 10 } = 14 \mathrm {~m} / \mathrm { s }$

Velocity with which it rebounds.

$v = \sqrt { 2 g h _ { 2 } } = \sqrt { 2 \times 9.8 \times 2.5 } = 7 \mathrm {~m} / \mathrm { s }$

$\therefore$ Change in velocity

$\Delta v = 7 - ( - 14 ) = 21 \mathrm {~m} / \mathrm { s }$

$\therefore$ Acceleration $= \frac { \Delta v } { \Delta t } = \frac { 21 } { 0.01 } = 2100 \mathrm {~m} / \mathrm { s } ^ { 2 }$ (upwards)