An aircraft flying horizontally at a height of 2Km with a speed 200m/s passes directly overhead an anti aircraft gun. At what angle from the vertical should the gun be fired so that a shell with muzzle speed 600 m/s may hit the plane? (Take g=10 m/s)

We're asked to find the necessary launch angle (with respect to the vertical) of the anti-aircraft shell to hit an incoming plane flying overhead.

I'll assume the shell launcher will fire as soon as the plane is directly overhead.

The position component equations for the plane are given by

• #x_"plane" = (200color(white)(l)"m/s")t#

• #y_"plane" = 2000color(white)(l)"m"#

And for the anti-aircraft shell:

• #x_"shell" = v_0cosalpha_0t#

• #y_"shell" = v_0sinalpha_0t - 1/2g t^2#

Ideally, the #x#-velocity for the plane and shell must be equal, because the #x#-component does not change; if they were different, it would never hit the target.

With that being said, the two quantities

#(200color(white)(l)"m/s")t#

and

#v_0cosalpha_0t#

should be the same; i.e.

#v_0cosalpha_0 = 200color(white)(l)"m/s"#:

The shell's initial velocity is #600# #"m/s"#:

#(600color(white)(l)"m/s")cosalpha_0 = 200color(white)(l)"m/s"#

#color(red)(alpha_0) = arccos[(200cancel("m/s"))/(600cancel("m/s"))] = color(red)(ul(70.5^"o"#

However, this is the angle with respect to the horizontal. The angle measured from the vertical is

#color(blue)("angle") = 90^"o" - color(red)(70.5^"o") = color(blue)(ulbar(|stackrel(" ")(" "19.5^"o"" ")|)#

We can even use this value to find the time it takes the projectile to hit the target, knowing that the height of the plane (#2000# #"m"#) should be set equal to the #y#-position equation for the shell:

#2000color(white)(l)"m" = (600color(white)(l)"m/s")sin[70.5^"o"]t - 1/2g t^2#

#2000color(white)(l)"m" = 593t - 5t^2#

#color(darkblue)(t = 3.48color(white)(l)"s"#

(Keep in mind wind resistance may affect this severely, so don't try and use this for a real-life scenario, as things could go terribly wrong).