Posted by **Liza** on Saturday, June 23, 2012 at 9:41am.

An object is released from rest at height h above the surface of the Earth, where h is much smaller than the radius of the Earth. It takes t seconds to fall to the ground. At what height should this object be released from rest in order to take 2t seconds to fall to the ground?

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Solution: We need to find the connection between the height and time. With “down” positive, so a=g is positive, and with v0 = 0, distance = (1/2) at^2 ? h = (1/2)gt^2 ? h = t^2. Therefore, if we double t, h must increase by a factor of 2^2 = 4.

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My Question: I don't understand how they are able to just equate h with t^2 (h=t^2). Then say that the answer is 4h. I thought the answer would be 2gh. But then why is it that gravity isn't included in the final answer?

THANKS!

- Physics -
**bobpursley**, Saturday, June 23, 2012 at 10:41am
It is in the final answer, but it has divided out.

h(t)=1/2 g t^2

newheight=1/2 g(2t)^2=1/2 4t^2

so to find newheight, divide the second equation by the first.

hewheight/oldheight=4

newheight=4*oldheight

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