A 1450 kg car goes over a vertical hill with a radius of 64.7 m. What is the maximum velocity that will keep the car on the road as it goes over the hill?

To find the maximum velocity that will keep the car on the road as it goes over the hill, we need to consider the forces acting on the car at the top of the hill.

First, we need to determine the net force acting on the car at the top of the hill. At the top of the hill, the car is momentarily in a state of free fall. The only force acting on the car is the force of gravity pulling down. This force can be calculated using the equation:

F = m * g

where F is the force of gravity, m is the mass of the car, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Next, we need to find the centripetal force required to keep the car moving in a circular path at the top of the hill. The centripetal force can be calculated using the equation:

F = m * v^2 / r

where F is the centripetal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the circular path.

At the top of the hill, the net force acting on the car is the difference between the gravitational force and the centripetal force:

Net Force = F_gravity - F_centrifugal

Since the net force must be directed towards the center of the circular path to keep the car on the road, the net force at the top of the hill should be positive (greater than zero).

Therefore, we can set up the following equation:

F_gravity - F_centrifugal > 0

Substituting the equations for F_gravity and F_centrifugal, we get:

m * g - m * v^2 / r > 0

Solving for velocity (v), we have:

v^2 < r * g

Taking the square root of both sides, we get:

v < sqrt(r * g)

Finally, we can calculate the maximum velocity that will keep the car on the road as it goes over the hill by substituting the given values:

v < sqrt(64.7 m * 9.8 m/s^2)

v < sqrt(633.06)

v < 25.2 m/s

Therefore, the maximum velocity that will keep the car on the road as it goes over the hill is 25.2 m/s.