Question

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?

Answer 1

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