A car moving across the bridge at point A has a speed of 72 km / h. What should be the radius

A car moving across the bridge at point A has a speed of 72 km / h. What should be the radius of curvature of the bridge for the driver at point A to be in a weightless state?

V = 72 km / h = 20 m / s.
P = 0.
g = 10 m / s2.
R -?
In a state of weightlessness, only gravity acts on the body. The body does not act on the support and does not stretch the suspension, that is, the body weight is 0: P = 0.
When passing the upper point of the convex bridge, 2 forces act on the car: gravity m * g directed vertically downward, force N of the bridge pressure directed vertically upward.
m * a = m * g + N – 2 Newton’s law in vector form.
For projections onto the vertical axis 2, Newton’s law will take the form: – m * a = – m * g + N.
N = m * g – m * a = m * (g – a).
The centripetal acceleration a is expressed by the formula: a = V ^ 2 / R.
N = m * (g – V ^ 2 / R).
According to Newton’s 3 Laws, the force N with which the bridge acts on the car is equal to the force P with which the car acts on the bridge: N = P.
P = m * (g – V ^ 2 / R) = 0.
g = V ^ 2 / R.
The radius of curvature R is expressed by the formula: R = V ^ 2 / g.
R = (20 m / s) 2/10 m / s2 = 40 m.
Answer: the radius of curvature of the bridge should be R = 40 m.