A particle is projected vertically from the ground takes time `t_(1)=1 s` upto point `A`, `t_(2)=3 s` from point `A` to `B`, and time `t_(3)=4 s` from point `B` to highest point. Find the height of the middle point of `A` and `B` from the ground.

Answer

`60m`

Explanation

Time taken bythe particle to reach the highest point
`=(t_(1)+t_(2))//2=4s`
At highes point, `0=u -g((t_(1)+t_(2)+t_(3))/(2))`
`0=u-10xx4 rArr u=40 m//s`
Time taken to reach upto midpoint of `A` and `B` ,
`t_(0)=t_(1)+((t_(2)-t_(1)))/(2) =2 s`
Height of midpoint
`h_(midde)=ut_(0)-(1)/(2) g t_(0)^(2)=40xx2 -(1)/(2) xx10xx2^(2)`
`=80-20=60 m`.