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))`

Options

A. `11 sec`
B. `6 sec`
C. `15 sec`
D. `5 sec`

Answer

A

Explanation

For particle-1 `y=sqrt(3)x-(gx^(2))/(2u^(2)(1//4))rArry=sqrt(3)x-(2gx^(2))/(u^(2))`
For particle-2 `y=x-(gx^(2))/(2u^(2)(1//2))rArry=x-(gx^(2))/(u^(2))`
`x-(gx^(2))/(u^(2))=sqrt(3)x-(2gx^(2))/(u^(2))`
`x(sqrt(3)-1)=(gx^(2))/(u^(2))rArr x=(u^(2))/(g) (sqrt(3)-1)`
for particle-1
`u(1//2)t_(1)=(gx^(2))/(g) (sqrt(3)-1)rArrt_(1)=(2u)/(g) (sqrt(3)-1)`
`u(1sqrt(2))t_(2)=(u^(2))/(g) (sqrt(3)-1)rArrt_(2)=(sqrt(2)u)/(g) (sqrt(3)-1)`
`Deltat=u//g (2-sqrt(2))(sqrt(3)-1)=10.9secapprox 11sec`.