Answer :
We start with the equation that gives the distance an object falls:
[tex]$$
s = 16t^2.
$$[/tex]
Since the branch is 90 ft above the soil, we set
[tex]$$
s = 90.
$$[/tex]
Plugging this into the equation gives:
[tex]$$
16t^2 = 90.
$$[/tex]
Next, solve for [tex]$t^2$[/tex]:
[tex]$$
t^2 = \frac{90}{16} = 5.625.
$$[/tex]
To find [tex]$t$[/tex], take the square root of both sides:
[tex]$$
t = \sqrt{5.625} \approx 2.3717.
$$[/tex]
Rounding to the nearest tenth of a second, we have:
[tex]$$
t \approx 2.4 \text{ seconds}.
$$[/tex]
Thus, it takes about [tex]$2.4$[/tex] seconds for the twig to reach the soil. This corresponds to option D.
[tex]$$
s = 16t^2.
$$[/tex]
Since the branch is 90 ft above the soil, we set
[tex]$$
s = 90.
$$[/tex]
Plugging this into the equation gives:
[tex]$$
16t^2 = 90.
$$[/tex]
Next, solve for [tex]$t^2$[/tex]:
[tex]$$
t^2 = \frac{90}{16} = 5.625.
$$[/tex]
To find [tex]$t$[/tex], take the square root of both sides:
[tex]$$
t = \sqrt{5.625} \approx 2.3717.
$$[/tex]
Rounding to the nearest tenth of a second, we have:
[tex]$$
t \approx 2.4 \text{ seconds}.
$$[/tex]
Thus, it takes about [tex]$2.4$[/tex] seconds for the twig to reach the soil. This corresponds to option D.
