Balancing the forces, we get
$M g-N=M \frac{v^2}{R}$
For weightlessness, $N=0$
$\therefore \frac{M v^2}{R}=M g$
where $R$ is the radius of curvature and $v$ is the speed of car.
Therefore, $v=\sqrt{R g}$
Putting the values, $R=20 \mathrm{~m}, g=10.0 \mathrm{~m} / \mathrm{s}^2$
So, $y=\sqrt{20 \times 10.0}=14.14 \mathrm{~m} / \mathrm{s}^2$
Thus, the speed of the car at the top of the hill is between $14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$.
Note: The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.