Question

Can someone please answer this math question

Answer 1

Answer: The second missile will destroy first missile at hieght of 64.76m above the ground at time 19.15475 seconds

Explanation:

Explanation:

The military is performing missile test,

The one missile is launched from the ground and its path is modeled by the equation as : y= f(x) =
`-4(x-10)^2 + 400`

After 10 seconds, Second heat-seeking missile is launched and its path is modeled by the equation as : y= g(x) =
`-2(x-20)^2 + 300`

After 10 seconds, first missile will be at height of f(10)

f(10) =
`-4(10-10)^2 + 400`

f(10) =
`400`

When both the missile collide each other at same hieght of y

Also. If x is time for f(x) then, time for g(x) will be x-10 such that Second heat-seeking missile is launched after 10 second of first missile is launched

At time of collision, f(x)=g(x-10)

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

`-4(x-10)^2 + 100 = -2(x-30)^2`

`2(x-10)^2 - 50 = (x-30)^2`

`2[(x)^2-20x+100]- 50 = (x)^2-60x+900`

`2(x)^2-40x+200 - 50 = (x)^2-60x+900`

`2(x)^2-40x = (x)^2-60x+750`

`(x)^2=-20x+750`

`(x)^2+20x-750=0`

Roots of x are 19.15475 and -39.1547

Negative time is not possible

Therefore, time at x=19.15475 seconds

And both missiles collide as y=f(19.15475 ) or g(19.15475 -10)

f(x) =
`-4(x-10)^2 + 400`

f(19.15475) =
`-4(19.15475-10)^2 + 400`

f(19.15475) = 64.76m above the ground.

Thus, The second missile will destroy first missile at hieght of 64.76m above the ground at time 19.15475 seconds