anonymous
  • anonymous
A person 6 ft tall stands 10 ft from point P directly beneath a lantern hanging 30 ft above the ground, as shown in the figure below. The lantern starts to fall, causing the person’s shadow to lengthen. Given that the lantern falls 16t2 ft in t seconds, how fast will the shadow be lengthening when t = 1?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I honestly wish this was an april fools joke....
amistre64
  • amistre64
it looks doable, you got the pic?
amistre64
  • amistre64
usually this is just a matter of relating similar triangles

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More answers

anonymous
  • anonymous
Sent it in a message cuz it gives away the school I'm at lol
anonymous
  • anonymous
yeah I set it up as 6/x=30/x+10
amistre64
  • amistre64
thats not a lantern; thats the sun lol
anonymous
  • anonymous
.....
anonymous
  • anonymous
lol
anonymous
  • anonymous
its the lantern you got me thinking for a sec
amistre64
  • amistre64
|dw:1333327872990:dw|
anonymous
  • anonymous
dL/dt would be 16 correct? but I don't get how to put a dL/dt in there lol
amistre64
  • amistre64
similar triangles does look like the key: 6 is to 30-d as s is to s+10 (s+10)/s = (30-d)/6 ; since we want s'; lets take the derivative and solve for s' but im gonna rewrite it 1+10/s = 5 -d/6 -10s'/s^2 = -d'/6 s' = d' s^2/60 we agree so far?
amistre64
  • amistre64
dL, as you call it; is 32t i believe
anonymous
  • anonymous
hmmm sec let me see if I get what you are saying
anonymous
  • anonymous
how did yo uget 1+10/s=5-d/6
amistre64
  • amistre64
ill splain that after i post this since i already typed it :) all this happens when t=1 soo im gonna go ahead and use the usual gravity version of this: -16t^2 + 30 at t = 1; 14 6 14 -- = ----- s s+10 6s+60 = 14s 60 = 8s s = 7.5
anonymous
  • anonymous
AH i got the part i just asked about lol
amistre64
  • amistre64
by comparing like sides of the similar triangles we get:\[\frac{6}{s}=\frac{30-d}{s+10}\] good
amistre64
  • amistre64
\[s' = d' \frac{s^2}{60}\] \[s' = d' \frac{(7.5)^2}{60}\] d = 16t^2; so d' = 32t at any given t; when t=1; d' = 32 \[s' = 32 \frac{(7.5)^2}{60}\] \[s' = 4 \frac{(7.5)^2}{7.5}\] \[s' = 4(7.5)\] maybe
anonymous
  • anonymous
This is way too hard for non-eng students to be doing lol thats the correct answer
anonymous
  • anonymous
you're too smart
amistre64
  • amistre64
yay!! lol
amistre64
  • amistre64
i am, ive been trying to dumb myself down by watching 3 and a half men tho so there is hope ;)
anonymous
  • anonymous
hahaha out of curiosity are you are grad student or math teacher or something? You always answer the most difficult questions like you are serving cake
amistre64
  • amistre64
im an undergraduate at the moment, working towards a BA in math so that i can go for the masters and beyond
anonymous
  • anonymous
You must be at harvard or an ivy or something you are a damn genius!
anonymous
  • anonymous
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amistre64
  • amistre64
:) well, i apparently did it right on this one but forgot how to do it by today
amistre64
  • amistre64
ahh, that 30-d part ....

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