Problema Solution
A missile is launched from the ground. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds.
After 1 second, the missile is 103 feet in the air; after 2 seconds, it is 192 feet in the air.
Find the height, in feet, of the missile after 5 seconds in the air
Answer provided by our tutors
Let
h(x) = ax^2 + bx + c, where a, b, and c are constants
Supposing the missile is at height 0 at time 0.
This means we have three points: (0, 0) (1, 103) and (2, 192)
h(0) = a*0^2 + b*0 + c
h(0) = 0 follows c = 0
h(x) = ax^2 + bx
h(1) = a*1^2 + b*1
h(1) = a + b
a + b = 103
h(2) = a*2^2 + b*2
4a + 2b = 192
We have the following system of equations:
a + b = 103
4a + 2b = 192
........
........
a = -7
b = 110
h(x) = -7x^2 + 110x
The height for a missile after 5 seconds in the air is:
h(5) = -7*5^2 + 110*5
h(5) = 375 ft
The height for a missile after 5 seconds in the air is 375 feet.