Question

A ball dropped on to the floor from a height of 10m rebounds to a height of 2.5m,if the ball is in contact with the floor for 0.02s, its average acceleration during that contact is

• A. 2100 $m/s^{-}2$
• B. 1050 $m/s^2$
• C. 4200 $m/s^{-}2$
• D. 9.8 $m/s^{-}2$

Answer 1

The correct option is B 1050 $m/s^2$

Ball hit groung at $n$ velocity

Reboumded at $v$ velocity

$\to x^2=2gH=2\times 9.8\times 10=196$

or $x=\sqrt{1}96=14m/sec$

now $O^2=V^2-2gh$

or $v^2=3\times 9.8\times 2.5=49$

or $r=\sqrt{4}9=7m/sec$

change in velocity = $14+7=21m/sec=0.02sec$

Average acceleration = $\frac{21}{0.02}=1050m/se{c}^2$