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Saul Waters

Feb 20, 2025

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, such that the top of the meter stick is 2.67 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.126 seconds to pass the length of the meter stick, how high above the ground was the acorn before it fell (assuming that the acorn didn't run into any branches or leaves on the way down)?

3 Answers

h = W²/(2t²g) - W/2 + t²g/8 = 1.0²/(2.1269.8) - 1.0/2 + .126²/8h = 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
Let d = height from where acorn fell.Let t₁= time when acorn reached the top of the meter stickLet t₂ = time when acorn reached the bottom of the meter stickWhen 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 get1 = 4.9(t₂² - t₁²)1 = 4.9(t₂- t₁)(t₂+ t₁)Given that (t₂- t₁) = 0.126 .....(3), we get1 = 4.9(0.126)(t₂+ t₁)t₂+ t₁= 1/(4.9 x 0.126)t₂+ t₁= 1.62 ......(4)Adding (3) and (4) we get2t₂= 1.746t₂= 0.873 secplug in (2) to get(d - 1.67) = (1/2)9.8(0.873)²d = 5.4 mThe acorn fell from 5.4 m height... Show More
classic free fall....y=0.5at^2ort=sqrt(2*y/a)where SQRT is square root and a= 9.8m/s^2suppose acorn falls from height ht1 is time when acorn is 2.67m above ground, that is wheny1=h-2.67mt2 is time when acorn is 1m lower or 1.67m above ground, that is wheny2=h-1.67m and we know that t2-t1=0.126secnow solve for h...sqrt(2(h-1.67)/9.81) - sqrt(2(h-2.67/9.81) = 0.126which 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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