Cut it!

Minimum condition will be satisfied, when the minimum condition for loop-a-loop will be satisfied. And as we know, this is possible only when the velocity at the bottom will be equal to the minimum velocity for loop-a-loop, i.e. $\sqrt { 5gr }$ , where $g$ is the acceleration due to gravity, and $r$ is the radius of the circle.

Now, by using energy conservation:

Energy at the bottom = Energy at the top

Using, ground as reference frame, Energy at bottom = $\frac { 1 }{ 2 } m{ (\sqrt { 5gr } })^{ 2 }$

Energy at the top = $mg(2r)+\frac { 1 }{ 2 } m{ v }^{ 2 }$ , where $2r$ is the body's height from the reference, and $v$ is it's velocity.

Since there are equal, solving the equation, $v$ comes out to be $\sqrt { gr }$ .

Now, after this point, the body will continue to go, and will follow the path of a horizontal projectile, whose height from the ground is $2r$ , initial velocity = $\sqrt { gr }$ , and whose Range is what we need to find.

Since in a horizontal projectile,

$Range(R)\quad =\quad u\sqrt { \frac { 2h }{ g } } \\ \Rightarrow \quad R\quad =\quad \sqrt { gr } \sqrt { \frac { 2(2r) }{ g } } \\ \Rightarrow \quad R\quad =\quad 2\sqrt { r } \sqrt { r } \\ \Rightarrow \quad R\quad =\quad 2r$

Centripetal force = $\frac{mv^2}{r}$ and the weight acting downwards is equal to ${mg}$ . These forces must be equal at the top of the loop so $\frac{mv^2}{r}$ = ${mg}$ $\Rightarrow$ $\frac{v^2}{r} = g \Rightarrow\ v= \sqrt{gr}$ . Now using distance = speed $\times$ time: $v^2 = \frac{d^2}{t^2} = \frac{(2r)^2}{t^2} \Rightarrow\ 4r^2 = ght^2 \Rightarrow\ t = \frac{2r}{\sqrt{gr}}$ Finally vertical displacement... $=\large vt = \sqrt{gr}\times\frac{2r}{\sqrt{gr}} = \boxed{2r}$

Alright listen up you Asian nimwhits. Prepare to get your minds blown and get taken on a ball wrenching trip to the land of intellectual bliss. I'm an American so you better listen up.

Firstly the change in x after coming out of this daggum circle is equal to its velocity tangent to the circle as it's initial x velocity and then its just a simple free fall problem.....Like the free fall India took when Britain made y'all their slaves.

so basically Velocity is 2 pi R over T and then when you multiply that by T your times cancel making it two.

Merica.