Question
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!
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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!
Answers
bobpursley
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
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