A jet plane files horizonally at a height h at a speed v. An anti-aircraft gun fires a shell at the plane when it is

vertically above the gun. Show that the minimum muzzle speed required to hit the plane is `sqrt(v^(2)+2gh)` at an angle `tan^(-1)sqrt(2gh)/(v)`

Explanation

Here , vt=`(u cos alpha)trArr v=u cos alpha`
`h=u sin at -(1)/(2)g t^(2)rArr(1)/(2)g t^(2)-(u sin alpha)t+h=0`
Since 't' is real, hence `u^(2)sin ^(2)alphage4xx(1)/(2)ghrArru^(2)(1-(v^(2))/(u^(2)))ge2gh`
`rArr u^(2)-v^(2)ge2gh rArr u gesqrt(v^(2)+2gh)`
`therefore u_("min")=sqrt(v^(2)+2gh)`
and `cos alpha=(v)/(u_("min"))=(v)/sqrt(v^(2)+2gh)rArr tan alphasqrt(2gh)/(v)`