A scientist dropped an object from a height of 200 feet. She recorded the height of the object in 0.5-second intervals. Her data is shown. Time (seconds): 0.0, 0.5, 1.0, 1.5, 2.0, 2.5; Height (feet): 200, 195, 185, 165, 135, 100 Based on a quadratic model, which best approximates the height at 3 seconds?
Question
Answer:
The quadratic model is given by:y = ax2 + bx + c
We evaluate three points to find the values of a, b and c.
We have then:
For (0, 200):
200 = a (0) 2 + b (0) + c
c = 200
For (0.5, 195):
195 = a * (0.5) ^ 2 + b * (0.5) +200
195-200 = a * (0.5) ^ 2 + b * (0.5)
-5 = 0.25a + 0.5b
For (1, 185):
185 = a * (1) ^ 2 + b * (1) +200
185-200 = a * (1) ^ 2 + b * (1)
-15 = a + b
Solving the system:
-5 = 0.25a + 0.5b
-15 = a + b
We have the following results:
a = -10
b = -5
Substituting:
y = -10x ^ 2 - 5x + 200
For x = 3 we have:
y = -10 (3) ^ 2 - 5 (3) + 200
y = - 90 - 15 + 200
y = 95
Answer:
the height at 3 seconds is:
y = 95 feet
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