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!
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
In the given problem, we are considering a scenario where the object is released from rest. This means that the initial velocity, v0, is zero. With this information, we can use the equation for distance, h, traveled by the object during free fall:
h = (1/2)gt^2
Here, g is the acceleration due to gravity, which is constant near the Earth's surface.
Now, we want to find the height at which the object should be released so that it takes 2t seconds to fall to the ground. Let's call this new height H. In this case, the equation for distance becomes:
H = (1/2)g(2t)^2
Simplifying this, we get:
H = (1/2)g(4t^2)
Now, let's compare the equation for H with the equation for h:
H = 4h
This tells us that the height H is four times the height h. So if an object takes t seconds to fall from height h, it will take 2t seconds to fall from a height that is four times larger. This is because the time it takes for an object to fall is directly proportional to the square root of the height.
Therefore, the answer is 4h, not 2gh, because we are comparing heights and not including the acceleration due to gravity in the final answer. The acceleration due to gravity (g) is already accounted for in the equations for h and H.
When analyzing the motion of an object in free fall near the surface of the Earth, we can use the equation h = (1/2)gt^2 to represent the height the object falls in terms of the time of fall. In this equation, h represents the height, g represents the acceleration due to gravity, and t represents the time of fall.
In the given problem, we are comparing two situations: the first situation in which it takes t seconds to fall to the ground from a certain height, and the second situation in which it takes 2t seconds to fall to the ground from a different height.
When we say that "h = t^2" for the first situation, we are essentially expressing the height in terms of the time. This equation is obtained by substituting the value of the acceleration due to gravity (g) from the equation (1/2)gt^2 because the distance traveled is the same as the height fallen.
Now, when we compare this equation to the second situation, where it takes 2t seconds to fall, we want to find the corresponding height. Since we established that h = t^2, if we double the time to 2t, the equation becomes h = (2t)^2 = 4t^2. This is where the result of 4h comes from.
The reason we don't explicitly include gravity (g) in the final answer is because the acceleration due to gravity remains constant throughout the free fall. It affects the time taken to fall, but when comparing the heights between the two situations, the acceleration cancels out.
In summary, the height at which the object should be released from rest in order to take 2t seconds to fall is 4 times the height at which it takes t seconds to fall. The factor of 4 comes from doubling the time and squaring it, which cancels out the acceleration due to gravity.