Question

a rock is launched from a canon. it's height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. After 1 second, the Rock is 58 ft in the air; after 2 seconds, it is 112 feet in the air. find the height, in feet, of the rock after 10 seconds in the air.

Answer 1

This is the quadratic function:
h(x)=ax²+bx+c

We have two points:
(1,58)
(2,112)

Now, we calculate this quadratic funtion.

we assume that h(0)=0
Therefore:
a(0)²+b(0)+c=0
c=0

(1,58)
a(1)²+b(1)=58
a+b=58  (1)

(2,112)
a(2)²+b(2)=112
4a+2b=112
2a+b=56  (1)

With the equations (1) and (2) we make a system of equations:
a+b=58
2a+b=56
we can solve this system of equations by reduction method.
-(a+b=58)
2a+b=56
---------------------
a=-2

-2(a+b=58)
 2a+b=56
-------------------
       -b=-60  ⇒  b=60

The function is:
h(x)=ax²+bx+c
h(x)=-2x²+60x

Now find the height, in feet, of the rock after 10 seconds in the air.

h(10)=-2(10)²+60(10)
h(10)=-200+600=400

Answer: 400 ft.

Answer 2

Answer: 400 ft.

For Plato