# Doubling the time spent in the air by a thrown object

• marksyncm
In summary: Simplifying the right hand side gives##H = \frac{1}{8} g t^2##In summary, to double the time spent in the air for an object thrown straight up from the ground and reaching a maximum height of H, we need to quadruple the initial height to H in order to get a total time spent in the air of 2t. This can be understood by rearranging the equation ##H = vt_a + \frac{1}{2} gt_a^2## to get ##H = \frac{1}{8} gt^2##, which shows that the initial height must be quadrupled to double the time spent in the air.
marksyncm

## Homework Statement

An object is thrown straight up from the ground and reaches a maximum height of H. How high up do we need to throw the object to double the time it spends in the air? Assume no air friction.

## Homework Equations

##\Delta x = vt+\frac{g}{2}t^2##

## The Attempt at a Solution

I solved this as follows:

1) Setting ##v=0## in the equation above, the time it takes the object to reach maximum height is ##t=\sqrt{\frac{2H}{g}}##
2) Total time spent in the air is equal to double the above, so to double the time spent in the air I just need to double the time ##t## it takes to reach peak height: ##2t = 2\sqrt{\frac{2H}{g}} = \sqrt{\frac{8H}{g}}##
3) Therefore, we quadrupled the height to double the time spent in the air.

What I do not understand is the solution presented to this problem in my textbook (Resnick/Halliday). Here's a screenshot:

I do not understand how we can go straight from the part encircled in red to the part encircled in black. I can see how ##H=\frac{1}{8}gt^2## follows from ##t=2\sqrt{\frac{2H}{g}}##, but I do not see how it follows from ##H=vt_a+\frac{1}{2}gt_a^2## simply based on the fact that the time is being doubled. Is there a step missing, or am I missing something?

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Compare the right hand side of both of those two equations.

Rearrange the right hand side of the second to get the left hand side.

marksyncm
Ahhh, I see it. Thanks.

Since ##t## is double ##t_a##, if the form ##\frac{1}{2} g t^2## is to give the same value H, then ##t## must be divided by two. That is,

##H = \frac{1}{2} g t_a^2 = \frac{1}{2} g \left( \frac{t}{2} \right)^2##

marksyncm

## 1. How does doubling the time spent in the air affect the distance traveled by a thrown object?

Doubling the time spent in the air will also double the distance traveled by a thrown object, assuming all other factors such as velocity and angle remain constant. This is because the distance traveled is directly proportional to the time spent in the air.

## 2. Is there a limit to how much the time spent in the air can be doubled for a thrown object?

Technically, there is no limit to how much the time spent in the air can be doubled for a thrown object. However, factors such as air resistance and gravity may affect the object's trajectory and limit the actual distance traveled.

## 3. How does the weight of a thrown object affect its time spent in the air?

The weight of a thrown object does not directly affect its time spent in the air. However, a heavier object may require more force to be thrown with the same velocity, which can affect the time spent in the air and the distance traveled.

## 4. Can the time spent in the air be increased by changing the angle of a thrown object?

Yes, the time spent in the air can be increased by changing the angle of a thrown object. A higher angle will result in a longer flight time, while a lower angle will result in a shorter flight time.

## 5. How does air resistance impact the time spent in the air for a thrown object?

Air resistance can significantly affect the time spent in the air for a thrown object. The force of air resistance acts on the object in the opposite direction of its motion, slowing it down and shortening its flight time. Objects with less aerodynamic shapes will experience more air resistance and have a shorter flight time compared to more streamlined objects.

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