A cannon fires successively two shells from the same point with velocity `V_(0)=250m//s`, the first at the angle `theta_(1)=60^(@)` and the second at the angle `theta_(2)=45^(@)` to the horizontal, the azimuth being the same. Neglecting the air drag, find the approximate time interval between firings leading to the collision of the shells `(g=9.8m//s^(2))`

Explanation

Let the first shell travel for time `t_(1)` and second fortime `t_(2)` before they collide.
Both the shells must travel equal horizontal and vertical distances to collide.
For horizontal motion, `v_(0)cos theta_(1).t_(1)=v_(0)cos theta_(2).t_(2)`
i.e. `(t_(1))/(t_(2))=(cos theta_(2))/(cos theta_(1)),(t_(1))/(t_(2))=(cos 45^(@))/(cos 60^(@))=(1//sqrt(2))/(1//2)t_(2)`
i.e. `t_(1)=sqrt(2)t_(2)`
For vertical motion
`v_(0)sin theta_(1).t_(1)-1/2"gt"_(1)^(2)=v_(2)sin theta_(2). t_(2)-1/2 "gt"_(2)^(2)` (or)`v_(0)sin 60^(@) . t_(1)-1/2"gt"_(1)^(2)=v_(0)sin 45^(@). t_(2)-1/2"gt"_(2)^(2)`
or `v_(0)(sin 60^(@)t_(1)-sin 45^(@)t_(2))=t/2(t_(1)^(2)-t_(2)^(2))`
Putting values `250-((sqrt(3))/2sqrt(2)t_(2)-1/(sqrt(2))t_(2))=9.8/2(2t_(2)^(2)-t_(2)^(2))implies250((sqrt(6)-sqrt(2))/2)t_(2)=4.9t_(2)^(2)`
`t_(2)=(250(2.45-1.41))/(2xx4.9)` or `t_(2)=(250xx1.04)/9.8=26.7simpliest_(1)=sqrt(2)t_(2)=1.414xx26.7=37.8s`
Time interval `t_(1)-t_(2)=37.8s-26.7s=11.1s`