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

the height, in feet, of a free-falling object t seconds after being dropped from an initial height s can be found using h=-16t^2+s. An object is dropped from a helicopter 784 feet above the ground. How long will it take the object to land on the ground?

how do you know where to put what??

a. 7s

b. 9s

c. 6s

d. 8.5s

- math -
**tchrwill**, Monday, February 22, 2010 at 2:49pm
h = 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 8:56pm
ok 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 10:34am
ok 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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