Question 372684: If an object is projected vertically upward from an altitude of s(o) feet with an intial velocity of v(o) ft/sec, then its distance above the ground after t seconds is: s(t)=-16t^2+ v(o)t+ s(o).
If s(1)= 64 and s(2)=66, what are s(o) and v(o)?
I honestly do not know how to set the problem up! Please help, it would be greatly appreciated!Thanks :)
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Given:
s(t)=-16t^2+ v(o)t+ s(o).
If s(1)= 64 and s(2)=66, what are s(o) and v(o)?
.
So,
when x=1 then s(1)=64
and
when x=2 then s(2)=66
.
From here, you now have two unknowns and two equations:
when x=1 then s(1)=64
s(t)=-16t^2+ v(o)t+ s(o)
s(1)=-16(1)^2+ v(o)(1)+ s(o)
64=-16+ v(o)+ s(o)
89=v(o)+ s(o) equation 1
.
when x=2 then s(2)=66
s(t)=-16t^2+ v(o)t+ s(o)
s(2)=-16(2)^2+ v(o)(2)+ s(o)
66=-16(4)+ 2v(o)+ s(o)
130=2v(o)+ s(o) equation 2
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Considering the "system of equations" then:
130=2v(o)+ s(o)
89=v(o)+ s(o)
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Multiply both sides of the bottom equation by -1 and add:
130=2v(o)+ s(o)
-89=-v(o)- s(o)
---------------------
41 ft/sec = v(0)
.
Substitute the above into:
89=v(o)+ s(o)
89=41+ s(o)
48 feet = s(o)
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