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November 28, 2014

November 28, 2014

Posted by **rmz** on Monday, February 22, 2010 at 1:59pm.

how do you know where to put what??

a. 7s

b. 9s

c. 6s

d. 8.5s

- math -
**tchrwill**, Monday, February 22, 2010 at 2:49pmh = Vo(t) + g(t^2) where h = the initial height, 784 feet, Vo = the initial velocity = 0, g = the acceleration due to gravity, 32 fps^2 and t = the time to fall from h = 784 to h = 0.

Therefore, 784 = (0)t + 32(t^2).

Solve for t.

- math -
**rmz**, Monday, February 22, 2010 at 8:56pmok I still do not understand this is what I got so far from the word problem above.

784=-16(0t^2)+s

784=-16+s

784+16=t^2+s

800

and the square root of 800 is 28

am I doing this right?

- math -
**tchrwill**, Tuesday, February 23, 2010 at 10:34amok I still do not understand this is what I got so far from the word problem above.

784=-16(0t^2)+s

784=-16+s

784+16=t^2+s

800

and the square root of 800 is 28

am I doing this right?

Using your terminology:

s = the altitude of the helicopter

t = the time from the release of the object from the helicopter

h = the height of the object above the ground after t seconds

h = -16(t^2) + s where -16 is the coefficient of t^2

You have altered the real equation to read h = -16 + t^2 + s, and then ignored the term "s".

Remember that s = the altitude of the helicopter = 784 feet and h = the height of the object (the ground)after "t" seconds or h = "0".

Then 0 = -16(t^2) + 784 or

16(t^2) = 784 or t^2 = 49 making t = 7 seconds.

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