SOLUTION: The military is performing missile tests. A new missile is launched from the ground and it’s path can be modelled by the quadratic equation y=-4(X-10)^2 +400 , where y is the mi

Algebra ->  Graphs -> SOLUTION: The military is performing missile tests. A new missile is launched from the ground and it’s path can be modelled by the quadratic equation y=-4(X-10)^2 +400 , where y is the mi      Log On


   



Question 1195454: The military is performing missile tests. A new missile is launched from the ground and it’s path can be modelled by the quadratic equation
y=-4(X-10)^2 +400 , where y is the missile’s height in meters and x is the time in seconds. 10 seconds later, a second heat-seeking missile is
launched to intercept and explode the first missile before it reaches the ground. The second missile’s path can be modelled by the quadratic equation
y= -2(x-20)^2+300
, where y is its height in meters and x is its time in seconds. At what time and height should the second missile
reach and explode the first missile?

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

y=-4%28x-10%29%5E2+%2B400......eq.1
y=+-2%28x-20%29%5E2%2B300.....eq.2
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to find at what time and height should the second missile reach and explode the first missile, find intersection point of the system above
-4%28x-10%29%5E2+%2B400=+-2%28x-20%29%5E2%2B300}......solve for x
-4x%5E2+%2B+80x+-400+%2B400=+-2x%5E2+%2B+80x+-800%2B300}
-4x%5E2+%2B+80x+=+-2x%5E2+%2B+80x+-500}....simplify
-4x%5E2+=+-2x%5E2+-500}
500+=+4x%5E2-2x%5E2+}
500+=+2x%5E2+}
250+=+x%5E2+}
x=sqrt%28250%29
x+=+5sqrt%2810%29->exact solution
x=15.8->approximately
find y

y=-4%285sqrt%2810%29-10%29%5E2+%2B400......eq.1
y=200%282sqrt%2810%29+-+5%29 -> exact solution
y=264.9 ->approximately
At time 15.8 seconds and height 264.9 meters the second missile should reach and explode the first missile