A fighter plane flying horizontally at an altitude of ` 1.5 km` with speed ` 720 kmh^(-1)` passes directly over head an anticraft gun.
At what angle from the vertical should the gun be fired from the shell with muzzle speed ` 600 ms^(-1)` to hit plane.
At what minimum altitude should the pilot fly the plane to avoid being hit ? ( Take g= 10 `ms^(-2)`).

Options

A. `sin^(-1)((1)/(3))` , 16km
B. `sin^(-1)((2)/(3))` ,16 km
C. `cos^(-1)((1)/(3))` 16m
D. `cos^(-1)((2)/(3))` ,16m

Answer

A

Explanation

Here,
Speed of the plane, `v=720kmh^(-1)=720xx(5)/(18)ms^(-1)`
`=200ms^(-1)`
Speed of the shell, `u=600ms^(-1)`
Let the shell hit the plane at L after time t if fired at an angle `theta` with the vertical from O.
For hitting, horizontal distance travelled by the plane = horizontal distance travelled by the shell. i.e., `vxxtusingthetaxxt`
`sintheta=(v)/(u)=(200ms^(-1))/(600ms^(-1))=(1)/(3)`
`theta=sin^(-1)((1)/(3))`