Kinematics With Portals

Let $t_{0A}$ be the time it takes for the ball to fall from its starting position to portal A, and let $t_{BA}$ be the time it takes for the ball to fall from portal B to portal A.

Also, let $v_{Ax}$ and $v_{Ay}$ be the x- and y- components of the ball's velocity just as it enters portal A. Let $v_{Bx}$ and $v_{By}$ be the x- and y- components of the ball's velocity just as it leaves portal B.

Note that because portal B is perpendicular to portal A, the x- and y- coordinates of the ball's velocity will simply "trade places." That is,

$v_{Bx} = v_{Ay}$ and $v_{By} = v_{Ax}$

First, we solve the kinematics equations describing the ball's trajectory from its starting point to portal A for $R$ and $v_{Ay}$ . Note that $v_{Ax} = v_{0}$ .

$R = v_{0} t_{0A}$

$H = \frac{1}{2} g t_{0A} ^{2}$ => $t_{0A} = \sqrt{\frac{2H}{g}}$

=> $R = v_{0} \sqrt{\frac{2H}{g}}$

$v_{Ay} = g t_{0A} = \sqrt{2gH}$

After the ball goes through the portal, the x- and y- coordinates "trade places," so:

$v_{Bx} = \sqrt{2gH}$ and $v_{By} = v_{0}$

Again, we use kinematics, but this time the ball's horizontal displacement is $\frac{R}{2}$ and its vertical displacement is $h$ , so

$\frac{R}{2} = v_{Bx} t_{BA}$

Substituting in our formulas for $R$ and $v_{Bx}$ , this becomes

$\frac{v_{0}}{2} \sqrt{\frac{2H}{g}} = t_{BA} \sqrt{2gH}$ => $t_{BA} = \frac{v_{0}}{2g}$

We also know that:

$h = v_{By} t_{BA} + \frac{1}{2} g t_{BA}^{2}$

Performing the appropriate substitutions, this becomes:

$h = \frac{v_{0}^{2}}{2g} + \frac{g}{2} \frac{v_{0}^{2}}{4g^{2}}$

which simplifies to:

$h = \frac{5v_{0}^{2}}{8g}$

Finally, plug in the numbes:

$h = \frac{5(2.53 \frac{m}{s})^{2}}{8(9.81 \frac{m}{s^{2}})} = \boxed{h = 0.408m}$