A rubber ball is dropped from a height of `5m` on a plane, where the acceleration due to gravity is not shown. On bouncing it rises to `1.8 m.` The ball loses its velocity on bouncing by a factor of

Options

A. `(16)/(25)`
B. `(2)/(5)`
C. `(3)/(5)`
D. `(9)/(25)`

Answer

B

Explanation

Key Idea: According to conservation of energy, potential energy of height h=kinetic energy at ground.
Potential energy=Kinetic energy
`ie, mgh=(1)/(2)mv^(2)`
`Rightarrow v=sqrt(2gh)`
If `h_(1) and h_(2)` are intial and final heights, then
`v_(1)=sqrt(2gh_(1)),v_(2)=sqrt(2gh_(2))`
Loss in velocity
`Deltav=v_(1)-v_(2)=sqrt(2gh_(1))-sqrt(2gh_(2))`
`therefore` fractional loss in velocity `=(Deltav)/(v_(1))`
`=(sqrt(2gh_(1))-sqrt(2gh_(2)))/(sqrt(2gh_(1)))`
`=1-sqrt((h_(2))/(h_(1)))`
Substituting the values, we have
`therefore (Deltav)/(v_(1))=1-sqrt((1.8)/(5))`
`=1-sqrt(0.36)=1-0.6`
`=0.4=(2)/(5)`