In this problem:
A car of mass $430\ \mathrm{kg}$ travels around a flat, circular race track of radius $178\ \mathrm{m}$. The coefficient of static friction between the wheels and the track is $0.266$.
The same car now travels on a straight track and goes over a hill with radius $178\ \mathrm{m}$ at the top.
What is the maximum speed that the car can go over the hill without leaving the road?
Correct answer: $41.766\ \mathrm{m/s}$.
Explanation:
$$\frac{mv^2}{r} = mg - N$$ where $N$ is the normal force acting on the car from the ground. The car will fly off the ground just when $N = 0$ so the maximum speed allowed will be $$\begin{align}v_{\text{max}} &= \sqrt{gr} \\ &= \sqrt{(9.8\ \mathrm{m/s^2})(178\ \mathrm{m})} \\ &= 41.766\ \mathrm{m/s}.\end{align}$$
The car is not driving upside down! So why is the force of gravity positive and the normal force considered negative in this problem?