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 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:

To solve this, we analyze each option using the line of best fit \( h = -21.962x + 114.655 \) and the table data:

Option 1: Time to hit the ground

• Actual time to hit ground: \( x = 4.6 \) sec (when \( h = 0 \)).
• For the line of best fit, set \( h = 0 \):

\( 0 = -21.962x + 114.655 \)
Solve for \( x \):
\( 21.962x = 114.655 \)
\( x = \frac{114.655}{21.962} \approx 5.22 \) sec.

• Difference: \( 5.22 - 4.6 = 0.62 \approx 0.6 \) sec.

This means the line of best fit predicts the object hits the ground ~0.6 seconds later than actual. This matches the first option.

Option 2: Initial height (when \( x = 0 \))

• Actual initial height: \( h = 100 \) m (from table, \( x = 0 \)).
• Line of best fit at \( x = 0 \): \( h = -21.962(0) + 114.655 = 114.655 \) m.

The line predicts a higher initial height, not lower. So this is incorrect.

Option 3: Height at \( x = 3.5 \) sec

• Actual height at \( x = 3.5 \): \( h = 40.0 \) m (from table).
• Line of best fit at \( x = 3.5 \):

\( h = -21.962(3.5) + 114.655 \)
\( h = -76.867 + 114.655 = 37.788 \approx 37.8 \) m.
The line predicts ~37.8 m, but actual is 40 m. So it does not correctly predict 40 m at 3.5 sec. Incorrect.

Option 4: Height difference for all times

Check a few times:

• At \( x = 0 \): Line predicts \( 114.655 \), actual \( 100 \) → difference \( 14.655 \) (not 4).
• At \( x = 0.5 \): Line: \( h = -21.962(0.5) + 114.655 = 103.724 \); actual \( 98.8 \) → difference \( 4.924 \) (close to 4, but not exact for all).
• At \( x = 1.0 \): Line: \( h = -21.962(1) + 114.655 = 92.693 \); actual \( 95.1 \) → difference \( -2.407 \) (negative, so line predicts lower).

Thus, the difference is not always 4 m greater. Incorrect.

Answer:

A. 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.