A cannon files successively two shells 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 time interval between firings leading to the collision of the shells.

Answer

`Deltat=(2v_0)/(g)(sin(theta_1-theta_2))/(costheta_1+costheta_2)=11s`.

Explanation

Let the shells collide at the point `P(x,y)`. If the first shell takes `t s` to collide with second and `Deltat` be the time interval between the fringes, then
`x=v_0cos theta_1t=v_0cos theta_2(t-Deltat)` (1)
and `y=v_0sin theta_1 t-1/2g t^2`
`=v_0sin theta_2(t-Deltat)-1/2g (t-Deltat)^2` (2)
From Eq. (1) `t=(Deltat cos theta_2)/(cos theta_2-cos theta_1)` (3)
From Eqs. (2) and (3)
`Deltat=(2v_0sin (theta_1-theta_2))/(g (cos theta_2+cos theta_1))` as `Delta t!=0`