Question

the table shows the height, in meters, of an object that is dropped as time passes until the object hits the ground.
falling object
time (sec),
x
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.6
height (m),
h
100
98.8
95.1
89.0
80.4
69.4
55.9
40.0
21.6
0
a line of best fit for the data is represented by ( h=-21.962x + 114.655 ). which statement compares the line of best fit
with the actual data given by the table?
according to the line of best fit, the object would have hit the ground 0.6 seconds later than the actual time the object
hit the ground.
according to the line of best fit, the object was dropped from a lower height.
the line of best fit correctly predicts that the object reaches a height of 40 meters after 3.5 seconds.
the line of best fit predicts a height of 4 meters greater than the actual height for any time given in the table.

Explanation:

Step1: Find the time when \(h = 0\) for the line of best fit

Set \(h=0\) in the equation \(h=-21.962x + 114.655\).

$$
LATEXBLOCK0
$$

The actual time when \(h = 0\) is \(x = 4.6\). The difference is \(5.2-4.6 = 0.6\)

Step2: Check the initial height (\(x = 0\)) for the line of best fit

When \(x = 0\), \(h=-21.962\times0+114.655 = 114.655\). The actual initial height is \(100\), so the object was dropped from a higher height according to the line of best - fit.

Step3: Check the height at \(x = 3.5\) for the line of best fit

When \(x = 3.5\), \(h=-21.962\times3.5+114.655=-76.867 + 114.655=37.788
eq40\)

Step4: Check the general height prediction

Take \(x = 0\): line of best - fit \(h = 114.655\), actual \(h = 100\), difference \(114.655 - 100=14.655
eq4\)

Answer:

According to the line of best fit, the object would have hit the ground \(0.6\) seconds later than the actual time the object hit the ground.