Two particles are simultaneously thrown from the top of two towards as shown. Their velocities are `2ms^(-1) and 14ms^(-1)`. Horizontal and vertical separations between these particles are 22 m and 9 m respectively. Then the minimum separation between the particles in the process of their motion in meters is `(g=10ms^(-2))`

Answer

`6`

Explanation

`v_(x)=8sqrt(2)s^(-1)=` relative velocity
along x-axis
`:.x=(22-8sqrt(2)t)`
`v_(y)=6sqrt(2)m//s=` relative velcoity along the y-axis
`:.y=(9-6sqrt(2)t)`
`r=sqrt(x^(2)+y^(2))`
For minimum r,
`(dr)/(dt)=0impliest=23/(10sqrt(2))s`
`r=sqrt(x^(2)+y^(2))`
Substitutingk the value of t in r,
`r_("min")=6m`