Question 1195069
y=-4(x-10)^2+400
y=2(x-20)^2+300 -> should be {{{y= -2(x-20)^2+300}}}

solve the system

{{{y=-4(x-10)^2+400}}}
{{{y= -2(x-20)^2+300}}}

The first and second missile will intersect at {{{x}}} that makes

{{{-2(x-20)^2+300=-4(x-10)^2+400}}}

{{{-2x^2 + 80x -500=80x - 4 x^2}}}..........divide by{{{ 2}}}
 {{{-x^2 + 40x -250=40x - 2 x^2}}}
{{{2 x^2-x^2 + 40x -250-40x =0}}}
{{{x^2  -250 =0}}}
{{{x^2  =250 }}}
{{{x =sqrt(250 )}}}
{{{x =15.81}}}

{{{y=-4(15.81-10)^2+400=264.98}}}

the point of intersection is ({{{15.81}}},{{{264.98}}})

so, after {{{15.81}}} seconds the second missile will reach and explode the first missile at  height of {{{ 264.98m}}}