A rock is launched from a cannon. After 1 second, the rock is 58 feet in the air; after 2 seconds, it is 112 feet in the air. Us ing the quadratic function, find the height, in feet, of the rock after 10 seconds in the air
1 answer:
Answer:
Step-by-step explanation: 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.
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