Question

A ball falls from 20 m height from floor and rebounds to 5 m . Time of contact is 0.02 s. Find the acceleration during the impact .

Answer 1

Let v1 be the velocity of the ball when dropped before striking the floor,

v2 be the velocity of ball upwards after striking the floor during rebounding

Change in velocity of ball during contact =v2 - (-v1)

Acceleration  $a = \frac{v_{2} - (-v_{1})}{\Delta t}$

Now from equation of motion ,

$v^2 = u^2 +2a \times s$

When the body dropped from height 20 m then,  $v=v_{1}$ , $u=0$ and $a= g =[tex]9.8m/s^2$

so, $v_{1}^2= 0+2\times 9.8 \times 20$

or

$v_{1} ^2 =392 m/s$

$v =19.8 m/s$

After striking the floor, in the equation of motion $u=v_{2}$ ,$v=0$ ,$a=-g$ and $s=5m$

Therefore,

 $0=v_{2}^2+2\times(-9.8 m/s^2)\times 5m$

or

$v_{2}^2=98$

$v=9.9 m/s$

Thus substituting these values in acceleration formula we get

$a=\frac{9.9-(-19.8)}{0.02 s}$

$a= 1485 m/s^2$