Respuesta :
We use the equation of motion,
[tex]S= ut+\frac{1}{2}at^{2}[/tex]
Here, S is the height, u is initial velocity and a is acceleration.
Given, [tex]S = 20 \ ft[/tex] [tex]S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft} = 6.096 \ m[/tex]
As acorn falls from tree, therefore we take the value of [tex]a = 9.8 \ m/s^2[/tex] and initial velocity [tex]u = 0[/tex].
Substituting these values in equation of motion,
[tex]6.096 \ m = 0 \times t +\frac{1}{2} \times 9.8 m/s^2 (t)^2 \\\\\ t = 1.12 \ s[/tex]
Thus, the time taken by the acorn to fall 20 feet ( 6.096 m ) is 1.12 s.
The time taken by acorn to touch the ground from 20 feet height is 1.12 s.
Second equation of motion
[tex]\rm \bold { S= ut + \frac{1}{2} at^2}[/tex]
Where ,
S- distance travelled = 20ft = 6.096 m
u- initial velocity = 0
t- time = ?
a- acceleration = [tex]\bold{ 9.8 m/s^2}[/tex]
Put the values to equation, we get
t = 1.12 s
Hence, we can say that the time taken by acorn to touch the ground from 20 feet height is 1.12 s.
To know more about Acceleration, refer to the link:
https://brainly.com/question/24372530?referrer=searchResults
