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?
Accepted Solution
A:
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