Answer:
[tex]\boxed {\boxed {\sf A. \ 2.7 \ m/s}}[/tex]
Explanation:
We want to find how fast a branch is falling after 0.28 seconds.
Essentially, we want to find its final velocity at exactly 0.28 seconds. We will use the following kinematic equation:
[tex]v_f= v_i+at[/tex]
The branch fell from the tree, so it initially started at rest or 0 meters per second. The branch is in free fall, so its acceleration is due to gravity, or 9.8 meters per second squared. It falls for 0.28 seconds.
- [tex]v_i[/tex]= 0 m/s
- a= 9.8 m/s²
- t= 0.28 s
Substitute the values into the formula.
[tex]v_f= 0 \ m/s + (9.8 \ m/s^2)(0.28 \ s)[/tex]
Multiply the numbers in parentheses.
[tex]v_f= 0 \ m/s +(9.8 \ m/s/s * 0.28 \ s )[/tex]
[tex]v_f= 0 \ m/s +2.744 \ m/s[/tex]
Add.
[tex]v_f= 2.744 \ m/s[/tex]
The original measurement of time has 2 significant figures, so our answer must have the same. For the number we found, that is the tenth place. The 4 in the hundredth place tells us to leave the 7.
[tex]v_f \approx 2.7 \ m/s[/tex]
The branch is moving at a velocity of approximately 2.7 meters per second.
