You attach a meter stick to an oak tree, such that the top of the meter stick is 2.67 meters above the ground?
You attach a meter stick to an oak tree, with the top of the meter stick reaching 2.67 meters above the ground. Later, an acorn falls from a point higher in the tree. If it takes the acorn 0.126 seconds to fall the length of the meter stick, how high above the ground was the acorn before it fell? Please assume that the acorn did not encounter any branches or leaves on its way down.
3 Answers
h = W²/(2t²g) - W/2 + t²g/8 = 1.0²/(2.1269.8) - 1.0/2 + .126²/8
h = 2.716 m where h is the distance from the acorn to the top of the stick. For height above ground, add 2.67 to this number:
H = h + 2.67 = 5.386 m... Show More
Oct 22, 2024
Let d = height from where acorn fell.
Let t₁= time when acorn reached the top of the meter stick
Let t₂ = time when acorn reached the bottom of the meter stick
When acorn is at the top of the meter stick
(d - 2.67) = (1/2)9.8t₁² ......(1)
When acorn is at the bottom of the meter stick
(d - 1.67) = (1/2)9.8t₂² ....(2)
Subtracting (2) from (1) we get
1 = 4.9(t₂² - t₁²)
1 = 4.9(t₂- t₁)(t₂+ t₁)
Given that (t₂- t₁) = 0.126 .....(3)
, we get
1 = 4.9(0.126)(t₂+ t₁)
t₂+ t₁= 1/(4.9 x 0.126)
t₂+ t₁= 1.62 ......(4)
Adding (3) and (4) we get
2t₂= 1.746
t₂= 0.873 sec
plug in (2) to get
(d - 1.67) = (1/2)9.8(0.873)²
d = 5.4 m
The acorn fell from 5.4 m height... Show More
classic free fall....
y=0.5at^2
or
t=sqrt(2*y/a)
where SQRT is square root and a= 9.8m/s^2
suppose acorn falls from height h
t1 is time when acorn is 2.67m above ground, that is when
y1=h-2.67m
t2 is time when acorn is 1m lower or 1.67m above ground, that is when
y2=h-1.67m
and we know that
t2-t1=0.126sec
now solve for h...
sqrt(2(h-1.67)/9.81) - sqrt(2(h-2.67/9.81) = 0.126
which you can do on paper but for not patient people, wolfram gives answer as 5.4m right away:
http://www.wolframalpha.com/input/?i=sqrt%282*%28x...... Show More
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