Question

A particle is dropped from some height. After falling through height h, the velocity of the particle becomes `v_(0)`. If it further falls through a distance `y(y lt lt h)`, find the approximate increase in velocity in terms of `v_(0),y` and `h`.

Answer 1

`O` to `A`: `v_(0)^(2)=2ghimplies v_(0)=sqrt(2gh)`
`O` to `B`: `v^(2)=2g(h+y)impliesv=sqrt(2g(h+y))`
Increase in velocity
`Deltav=v-v_(0)=sqrt(2g(h+y))-sqrt(2gh)`
`=sqrt(2gh)[(1+y/h)^(1//2)-1]`
`=v_(0)[1+1/2y/h-1]=(v_(0)y)/(2h)`
Recall: Binomial theorem
`[1+y/h]^(1//2)=1+1/2y/h (since y lt lt h)`