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. `4 sec`
B. `7 sec`
C. `17 sec`
D. `11 sec`

Answer

D

Explanation

At collision horizontal distance covered by the both the particles are equal
`250 1/2 t_(1)=250 1/sqrt2`….(1)
at collision vertical distances covered by the both the particles are equal
`250sqrt3/2t_(1)-1/2 "g t"_(1)^(2)`
`= 250 xx(1)/(sqrt(2))t_(2)-(1)/(2)"g t"_(2)^(2)`....(2) …(2)
solve (1) & (2) get `t_(1)t_(2)`