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	21.8	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: Analyze when object hits ground

The object hits the ground when \( h = 0 \). For the line of best fit \( h=-21.962x + 114.655 \), set \( h = 0 \):
\( 0=-21.962x + 114.655 \)
\( 21.962x=114.655 \)
\( x=\frac{114.655}{21.962}\approx5.22 \) (wait, no, actual time is 4.6 sec. Wait, recalculate: \( x=\frac{114.655}{21.962}\approx5.22 \)? No, wait, maybe miscalculation. Wait, \( 21.962\times4.6\approx22\times4.6 = 101.2 \), \( 21.962\times5.2\approx22\times5.2 = 114.4 \), close to 114.655. So \( x\approx\frac{114.655}{21.962}\approx5.22 \)? But actual time is 4.6 sec. Wait, no, the first option says "0.6 seconds later". 5.2 - 4.6 = 0.6? Wait, maybe I miscalculated. Wait, \( h = -21.962x + 114.655 \), set h=0: \( x=\frac{114.655}{21.962}\approx5.22 \)? No, 21.962*5 = 109.81, 114.655 - 109.81 = 4.845, 4.845/21.962≈0.22, so x≈5.22? But actual time is 4.6. 5.22 - 4.6≈0.62, close to 0.6. So first option: line of best fit time is ~5.22, actual is 4.6, so difference ~0.6, so line of best fit time is 0.6 later.

Step2: Analyze initial height (x=0)

For line of best fit, at x=0, \( h = 114.655 \), actual initial height is 100. So line of best fit says dropped from higher height, so second option is wrong.

Step3: Check x=3.5, h=40

For line of best fit, x=3.5: \( h=-21.962(3.5)+114.655=-76.867 + 114.655 = 37.788 \), actual h=40. So line of best fit predicts 37.788, actual 40, so not correct. Third option wrong.

Step4: Check height difference

Take x=0: line h=114.655, actual 100, difference ~14.655. x=0.5: line h=-21.962(0.5)+114.655=-10.981 + 114.655=103.674, actual 98.8, difference ~4.874. Not 4 meters for all. Fourth option wrong.

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.