Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with speed u at angle `theta` with the horizontal.

Answer

`5 m`

Explanation

Velocity time curve of the process is shown. Acceleration before applecation of brake,
.
`a_(1)=(v_(m))/(t_(1)) rArr t_(1)=v_(m)/(a_(1))` (i)
Displacement =`(1//2) v_(m)`
`(t_(1)+t_(2)//2)`
But velocity `(v_(m)//2)` is observed at `C` and `D`.
`CD=Delta=Delta t=1 s` (given)
`C` and `D` are mid-points of `OA` and `OB` .
From similar triangles, `OB =(t_(1)_t_(2)//2) =2CD =2Delta t`
Displacement=`((1)/(2)) v_(m) (2 Delta t)=v_(m) Delta t =5 xx1=5 m`,