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)/(cos theta_(2)) , (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_(0) 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)) = (g)/(2) (t_(1)^(2) -t_(2)^(2))`
Putting values `250((sqrt(3))/(2)sqrt(2)t_(2) - (1)/(sqrt(2))t_(2)) = (9.8)/(2)(2t_(2)^(2) - t_(2)^(2))rArr 250((sqrt(6) - sqrt(2))/(2)) t_(2)= 4.9 t_(2)^(2)`
`t_(2) = (250(2.45-1.41))/(2 xx 4.9) (or) t_(2) = (250xx 1.04)/(9.8) = 26.7 s rArr t_(1) = sqrt(2) t_(2) = 1.414 xx 26.7= 37.8s`
Time interval `t_(1) - t_(2) = 37.8s - 26.7 s = 11.1 s`