SOLUTION: If an object is projected vertically upward from an altitude of a feet with an initial velocity of v ft/sec, then its distance s(t) above the ground after t seconds is
s(t) = W
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Question 681777: If an object is projected vertically upward from an altitude of a feet with an initial velocity of v ft/sec, then its distance s(t) above the ground after t seconds is
s(t) = −16t2 + vt + a.
If s(1) = 99 and s(4) = 129, what are v and a?
Found 2 solutions by mananth, stanbon:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
s(t) = −16t2 + vt + a.
If s(1) = 99 and s(4) = 129, what are v and a?
99= -16(1)^2+v(1)+a
99=v+a-16
v+a=99+16
v+a=115..........(1)
129= -16(4^2)+4v+a
129=-256+4v+a
4v+a=129+256
4v+a=386...........(2)
Solve (1) & (2)
subtract (2) from (1)
-3v=-276
v=92
Plug v in equation above to get a
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
If an object is projected vertically upward from an altitude of a feet with an initial velocity of v ft/sec, then its distance s(t) above the ground after t seconds is
s(t) = −16t2 + vt + a.
If s(1) = 99 and s(4) = 129, what are v and a?
-------
Substitute and solve:
s(4) = -16*16 + 4v + a
129 = -256 + 4v + a
4v + a = 385
-----
s(1) = -16 + v + a
v + a = 99 + 16 = 115
--------------------------
You have 2 equations:
4v + a = 385
v + a = 115
----
Subtract and solve for "v":
3v = 270
v = 90
----
Solve for "a" using v + a = 115
90 + a = 115
a = 25
----------------
Cheers,
Stan H.
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