Doubling the time spent in the air by a thrown object

[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?

 

[CWatters]

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.

 

[gneill]

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##