MATH SOLVE

5 months ago

Q:
# 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

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