Reduce and clip voltage of a 5Mhz ultrasonic signal from +-50V to 0-3.3 V
The high voltage signal (+/- 50 Volts, i.e. 100 Volts peak to peak) can be attenuated and DC biased using a filter like this:



Note that all the components used are standard values. The output will be a signal of 3.242 Volts peak to peak, with a 1.65 Volts DC bias added. Thus, the resultant output to the ADC pin will not have any negative values, which addresses the concern mentioned in the question.

Using high precision (low tolerance %, e.g. 1%) tantalum capacitors and resistors, and keeping all traces / leads short, will prevent much variance in values and little stray inductance / capacitance.

Why this works:


The combination of the 15 KOhm resistor and 1 pF capacitor acts as the upper half of a voltage divider. Impedance Z1 of the R and C in series @ 5 MHz is 35188 Ohms 
The 27 pF capacitor acts as the lower half of the voltage divider. Impedance @ 5 Mhz is 1179 Ohms
The 1 pF capacitor also blocks any DC component of the input signal
Output of the voltage divider by Ohms Law: . 
For Vin = +/- 50 Volts = 100 Volts p-p, Vout = +/- 1.621 Volts = 3.242 Volts peak to peak
The two 1 MOhm resistors again form a voltage divider for the 3.3 Volts Vcc, resulting in a DC bias of 1.65 Volts
Adding this bias to the signal gives an output signal ranging from 0.029 Volts to 3.271 Volts for the -50 to +50 Volts input, with a small phase shift.


Output impedance of this filter is around 1.14 KOhms, and input impedance is around 36.36 kOhms.