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OpenStudy (kaylak12345):
OpenStudy (kaylak12345):
@sooobored
OpenStudy (kaylak12345):
OpenStudy (kaylak12345):
OpenStudy (kaylak12345):
1. is still 5/4 right
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OpenStudy (sooobored):
is there any additional information above question 1, that gives the height of the building?
OpenStudy (kaylak12345):
you haveto come up with a height of a building
OpenStudy (sooobored):
oh, then what height did you decide on?
OpenStudy (kaylak12345):
25
OpenStudy (kaylak12345):
ft
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OpenStudy (sooobored):
ok, from this point on h0 =25ft for the rest of the problem
then 1. yes 5/4 would be correct
OpenStudy (kaylak12345):
yay
OpenStudy (sooobored):
wont help you graph,
2. -16t^2
since its negative it opens downwards
16t^2 would open upwards
we already went over this
OpenStudy (sooobored):
you should remember this type of parabola behavior as its very useful in future problems
OpenStudy (sooobored):
ok 3.
\[h=-16t^2 +v_0 t + h_0\]
vo is given as 150 upwards so
vo=150
we defined h0 = 25 from your first problem
so plug those in and you're get your equation
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OpenStudy (kaylak12345):
-16t^2+150t+25?
OpenStudy (sooobored):
dont forget to put h= infront of it
but yes
OpenStudy (sooobored):
3b.
you set h=0 since h=0 is the height of the ground
then solve for t
\[0=-16t^2+150 t +25\]
OpenStudy (kaylak12345):
-16t^2 =-25?
150t=187.5
OpenStudy (sooobored):
uhhhh i have no idea what you just did but it doesnt look right
you need to factor the binomial
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OpenStudy (sooobored):
or use the quadratic formula to determine t
OpenStudy (kaylak12345):
9.54
OpenStudy (sooobored):
sorry, i honestly dont know since its not something i can do in my head
OpenStudy (kaylak12345):
oh well i did it in a quadratic calcultor thats what it gave me
OpenStudy (sooobored):
should be about 9.5388
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OpenStudy (sooobored):
so yes
OpenStudy (kaylak12345):
now i have to make a table and thats what it gave me but we usually cut off after 2 numbers
OpenStudy (sooobored):
you can make the table easily yourself, just pick t values and get the solution by "plugging in"
skipping to 3f
OpenStudy (sooobored):
why cant you have negative t and h(t) values?
OpenStudy (kaylak12345):
It wouldnt make sense because the point of fireworks is to go up and negative height wouldnt make any sense at all. Its how tall something is
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OpenStudy (sooobored):
on the right track, h=1 means 1 ft off the ground h=-1 ,means 1 ft under the ground
doesnt make a lot of sense
now what does a negative t mean?
OpenStudy (kaylak12345):
-t means less time you cant subtract time
OpenStudy (sooobored):
sorta of, it means going back in time from an event
OpenStudy (sooobored):
and time travel is not possible yet
OpenStudy (kaylak12345):
lol
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OpenStudy (sooobored):
ok quick overview of 4.
its asking you to change vo
so instead of using 150 like in problem 3, you can choose something like 80
OpenStudy (kaylak12345):
so any number
OpenStudy (kaylak12345):
ok 81 it is
OpenStudy (sooobored):
sorry, trying to think of a nice number that would work
OpenStudy (sooobored):
399 is a very nice number
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OpenStudy (kaylak12345):
so i have to pick a firework?
OpenStudy (sooobored):
4. is choosing another initial velocity, which was vo=150 ft/s in problem 3
OpenStudy (kaylak12345):
ok so 79 then so how would i set this up the same way in question 3
OpenStudy (sooobored):
no no use 399, its easier
OpenStudy (kaylak12345):
ok how would i set this up to solve?
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OpenStudy (sooobored):
h=-16t^2 +399 t +25
look familiar?
OpenStudy (kaylak12345):
ok and yes
OpenStudy (sooobored):
oh, actually 399 doesnt help as much since you dont need to calculate when it lands on the ground
OpenStudy (sooobored):
anyways, 4b is asking when t=3, what is the height?
so plug in 3
OpenStudy (sooobored):
are you familiar with how to find the symmetry point?
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OpenStudy (kaylak12345):
kind of
OpenStudy (sooobored):
ok quick run through then
OpenStudy (sooobored):
you're familiar with the quadratic formula
\[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
OpenStudy (kaylak12345):
yes
OpenStudy (sooobored):
we can divide that into 2 parts
\[\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}\]
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OpenStudy (kaylak12345):
144+1197+25 i think im off on the height for 4b
OpenStudy (sooobored):
now parabola are always symmetrical
so the center line "x" is always \[\frac{-b}{2a}\]
OpenStudy (sooobored):
and the other component of the quadratic equation, is the distance the roots are away from the symmetry point
OpenStudy (sooobored):
-144
OpenStudy (sooobored):
all other numbers are correct, so just add them all together
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OpenStudy (kaylak12345):
1222 for b
OpenStudy (sooobored):
no, 1197+25-144
OpenStudy (kaylak12345):
oh crap
OpenStudy (kaylak12345):
1078
OpenStudy (sooobored):
right
well hopefully you can do the rest on your own since i gotta run
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OpenStudy (kaylak12345):
ok thank you for the help
OpenStudy (sooobored):
though ive covered everything you need to finish the rest