The biggest problem with this question is the scale of that diagram, to give you an idea:

  - The earth is 63.7x the height of the atmosphere in radius.
  - This means that leaving the earth horizontally will encounter much more atmosphere than shown.

A plotting of a circular model of the earth with atmosphere/without using equations shows this:

[![Earth plus atmosphere.][1]][1]

[Wolfaram Alpha Scaled Plot Here](https://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3D(6371)%5E2+AND+x%5E2%2By%5E2%3D(6371%2B100)%5E2)

But, more importantly, we want the intercept of where you'll be horizontally leaving the atmosphere...

[![enter image description here][2]][2]

So, basically, you're going to choose to travel through 1133.22km of atmosphere instead of 100km. You're doing 11.3322x the work to do what you've said (simply put, not including massive complications from prolonging a bunch of other stress on the materials). The goal is to minimize stress on the materials, if you've chosen to prolong stress to any substance you're jeopardizing the mission. 

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Once again, I am an amateur, let me know if any of this is wrong. Note, this information is simplified:

  - Does not include earth rotation.
  - This assumes the earth is a sphere, not true.
  - As JCRM pointed out the atmosphere is a gradient, so the amount of total atmosphere passed through is significantly less than 11.3322x.
    - You ***are*** increasing time at max Q though, because the gradient is more spread out.

  [1]: https://i.sstatic.net/YABcv.png
  [2]: https://i.sstatic.net/YLiDp.png