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