If block A is moving horizontally with velocity `v_(A)`, then find the velocity of block B at the instant as shown in fig.

Explanation

(`a`) Position vectors of the block and the stone are
`vecr_(B)=(2t)hati+(40)hatj`
`vecr_(S)=v_(1)thati+v_(2)thatj+(20t-(1)/(2)"gt"^(2))hatk`
When the stone hits the block
`2t=v_(1) t` , `40 m=v_(2) t` , `20 t-(1)/(2)"gt"^(2)=0`
`v_(1)=2m//s` , `v_(2)=(40)/(t)m//s` , `t=4 s`
` rArr v_(2)=10m//s`.
(`b`) For the same speed of throw, the maximum range
`R_(max)=(v^(2))/(g)=(v_(1)^(2)+v_(2)^(2)+20^(2))/(g)=(504)/(10)=50.4m`
` rArr BC=30.7m`.
Time taken by the stone to reach `C=(30.7)/(2)=15.35 s`
Time of fligh' of the stone `=(2(v//sqrt(2)))/(g)=(sqrt(2)sqrt(504))/(10)=3.17s`
Maximum value of `t=15.35-3.17=12.18 s`.