There is a simpler method for getting a ballpark guess of what increased voltage will do.
Power = Torque x speed.
Compare the two voltages at 50% free speed. (that's 1/2 free speed x 1/2 stall torque).
Both the free speed and stall torque linearly increase with voltage.
So, if at V1, Torque = T, and speed = S, then at V2, torque = V2/V1T and speed = V2/V1S
for a switch from 7.2 to 8.4 V, this works out to a new power of 1.36 the original. This is at half free speed, but the ratio would be true anywhere on the curve.
This also shows that double voltage quadruples power, which is a pretty cool fact. Increasing voltage certainly does not increase power linearly like many people think,
Your calculation with velocity cubed is a bit rough because you don't get constant power out of the motors, and drag isn't the only factor contributing to top speed.
ORIGINAL: Argess
That was a better answer than I can give as you gave a number relating to performance.....5 mph increase. But here's what I would have said anyway:
Power goes up by the voltage squared: P=V^2/R where R is the resistance, or in simple terms, the motor.
So for say a motor provides 1/2 Hp (373 watts) with a 6 volt battery, using the above forumla (and this is rough as the whole motor/esc isn't quite as simple as a purely resistive load), we get R = 0.097 ohms
Using this figure and re-calculating for 7.2 volts, we get 537 watts, or 0.72 Hp.
That's a 44% increase in power.
That seems high, and it's probably because I treated the motor/ESC as a simple resistive load, but it's an indication.
But let's go further. We know for top speed, the HP required is proportional to the Velocity cubed.
Therfore, if your vehicle does 39 mph with the 6 volts battery, it will do 44 mph with a 7.2 volt battery. And that is a 5 mph increase.
If your vehicle only goes 20 mph with a 6 volts battery, it will now go 22.6 mph with a 7.2 volt battery. And that's a 2.6 mph increase.