operational-amplifiervoltage
Why does this result in a 15 V output voltage?
I played a little with different values such as 0.2 V for input, 10k ground resistor and feedback resistor of 80k and the formulas for voltage gain:
$$ V_{gain} = \frac {R_f}{R_{ground} +1 } $$
and
$$ V_o = gain \times V_{in} $$
and it worked but not with image settings.
Bringing the resistances out
Starting from your final equation, $$\frac{V_o}{V_i} = \frac{1}{\frac{R_1}{R_f}-j\frac{f_c}{f}}\tag{1}$$ where \$f_c =\dfrac{1}{2\pi R_f C}\$.
Let \$f_1 = \dfrac{1}{2\pi R_1 C}\$ then, $$f_c = \frac{R_1}{R_f}\times f_1$$ Applying this in \$(1)\$, $$\frac{V_o}{V_i} = \frac{1}{\frac{R_1}{R_f}-j\frac{f_1}{f}\times \frac{R_1}{R_f}}$$ $$\left|\frac{V_o}{V_i}\right| = \left|\frac{R_f}{R_1}\times\frac{1}{1-j\frac{f_1}{f}}\right|$$ $$= \frac{R_f}{R_1}\frac{1}{\sqrt{1+\frac{f_1^2}{f^2}}}$$
Alternate method
You should have started this way. Especially when you had the final answer with you. :)
$$ \frac {V_o}{V_i} = \frac {R_f}{R_1 - \frac {j}{\omega C}}$$ $$= \frac{R_f}{R_1}\frac{1}{1-j\frac{1}{\omega R_1 C}}$$ $$= \frac{R_f}{R_1}\frac{1}{1-j\frac{f_1}{f}}$$
Taking absolute value results in the required result.
Your hyperphysics link is correct, and so is your conclusion about the input impedance of the voltage follower. The math follows from the basic control system diagram. You can see a nice presentation on it here:
http://slideplayer.com/slide/1496732/
Best Answer
More than likely the model of the op-amp you chose has built-in output voltage limiting to +/- 15 volts hence, with a gain of 1 + 50/5 (=11) and an input of +3 volts, it cannot produce a realizable output voltage of 33 volts and hits the built-in end-stops of +15 volts.