SOLUTION: A balloonist throws a sandbag downward at 24 feet per second from an altitude of 720 feet. Its height (in feet) above the ground after t seconds is given by
S(t) = 16t2 - 24t + 7
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Question 322956: A balloonist throws a sandbag downward at 24 feet per second from an altitude of 720 feet. Its height (in feet) above the ground after t seconds is given by
S(t) = 16t2 - 24t + 720.
a) Find S(1).
b) What is the height of the sandbag 2 seconds after it is
thrown?
c) How long does it take for the sandbag to reach the
ground? [On the ground, S(t ) = 0.]
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You aren't going to have much luck using that function. Up is positive and down is negative. The acceleration due to gravity is ALWAYS down, unless you have invented some sort of anti-gravity machine. Hence:
\ =\ -16t^2\ -\ 24t\ +\ 720)
\ =\ -16(1)^2\ -\ 24(1)\ +\ 720)
Which is the height of the object 1 second after it is thrown. Hence the height 2 seconds after it was thrown is )
(you can do your own arithmetic in both cases.)
^2\ -\ 24(1)\ +\ 720\ =\ 0)
Factors:
(t\ -\ 6)\ =\ 0)
Exclude the negative root.
John
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