SOLUTION: Sorry this is a really long problem and I am completely stumped about how to start and what to do. Chuck is standing atop a high platform and fires a rock up into the air with h

Question 205704: Sorry this is a really long problem and I am completely stumped about how to start and what to do.
Chuck is standing atop a high platform and fires a rock up into the air with his slingshot. While it is in flight, the rock's distance about the ground is a quadratic function of time. 1 second after firing it, the rock is 68 meters high, at 2 seconds it's 96 meters, and at 3 seconds it's 114 meters above ground. Use this information to answer the following questions.
a. Find and write the quadratic equation for the function
b. What is the highest point the rock will be above ground?
c. How high is the platform from which Chuck fires the rock? (Hint: Time = 0)
d. When will the rock hit the ground?
e. On its way down, instead of hitting the ground, the rock goes into a nearly empty well. It splashes into the water (hits bottom) exactly 10 seconds after Chuck fired it. How deep is the well?
Like I said really long and really confusing for me. Any help is really appreciated
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Chuck is standing atop a high platform and fires a rock up into the air with his slingshot. While it is in flight, the rock's distance about the ground is a quadratic function of time. 1 second after firing it, the rock is 68 meters high, at 2 seconds it's 96 meters, and at 3 seconds it's 114 meters above ground. Use this information to answer the following questions.
a. Find and write the quadratic equation for the function
Because a "quadratic equation" is:
f(t) = at^2 + bt + c
We have three unknowns: a, b, and c
Therefore, we need three equations. We get this from:
1 second after firing it, the rock is 68 meters high,
at 2 seconds it's 96 meters, and
at 3 seconds it's 114 meters above ground.
.
Our three equations are:
68 = a(1)^2 + b(1) + c
96 = a(2)^2 + b(2) + c
114 = a(3)^2 + b(3) + c
.
Or, rewritten:
68 = a + b + c
96 = 4a + 2b + c
114 = 9a + 3b + c
.
Solve the above system of equations to find a, b and c. Then put it back into:
f(t) = at^2 + bt + c
This will be your "quadratic equation" describing the the height (meters) based on time (sec).
.
b. What is the highest point the rock will be above ground?
find the vertex of the quadratic found in parta
.
c. How high is the platform from which Chuck fires the rock? (Hint: Time = 0)
Set t=0 and solve. It will be 'c'.
.
d. When will the rock hit the ground?
Set f(x)=0 and solve for t
0 = at^2 + bt + c
.
e. On its way down, instead of hitting the ground, the rock goes into a nearly empty well. It splashes into the water (hits bottom) exactly 10 seconds after Chuck fired it. How deep is the well?
Add 10 the answer from part d and plug it into your quadratic equation to find f(x), it will be a negative number (because it is below the ground). The absolute value will be the depth of the well.
Write back if you get stuck.
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