A body is oscillating vertical about its mean position. If its maximum and minimum heights above the surface of the ground are 2 m and 1.5 m, then the speed at the mean position will be `(g=10m//s^(2))`

Options

A. `5 m//s`
B. `sqrt(2.5)m//s`
C. `10 m//s`
D. `sqrt(10) m//s`

Answer

B

Explanation

`L=h_("max")-h_("min")`
`=2-1.5`
`2A=0.5" "therefore A=0.25`
`v_(m)=Aomega`
`=0.25sqrt((g)/(A))`
`=sqrt((10xx0.25xx0.25)/(0.25))=sqrt(2.5)m//s`