Two projectiles are projected simultaneously from two towers as shwon in figure. If the projectiles collide in the air, then find the distance "s" between the towers.

Explanation

We see here that projectiles are approaching both horizontally and vertically. Their movement in two component directions should be synchronized so that they are at the same position at a particular given time. For collision, the necessary requirement is that relative velocity and displacement should be in the same direction. It is given that collision does occur. It means that two projectiles should cover the displacement with relative velocity in each of the component directions. In x-direction,
`v_(AB)=u_(Ax)-u_(Bx)=10sqrt2cos45^(@)-(-10)`
`=10sqrt2(1)/sqrt2+10=20m//s`

if "t" is time after which collision occurs, then
`impliess=v_(Ay)-u_(By)`
`impliesv_(ABy)=ucos45^(@)-0=10sqrt2xx1/(sqrt2)=10m//s`
The initial vertical distance between points of projection is 30 – 10 = 20 m. This vertical distance is covered with component of relative velocity in vertical direction. Hence, time taken to collide, "t", is :
`impliest=(20)/(10)=2`
Putting this value in the earlier equation for "s", we have :
`impliess=20t=20xx2=40m`