What is the maximum speed a car can travel over the top of a hill of radius 22 m before flying off the hill?

g=v^2/r

solve for v. g=9.8m/s^2

gravity provides the centripetal force

m g = m v^2 / r ... v^2 = r g = 22 * 9.8

answer is in m/s

To determine the maximum speed a car can travel over the top of a hill before flying off, we need to consider the gravitational force and the centrifugal force acting on the car.

First, let's calculate the gravitational force acting on the car. The gravitational force is equal to the weight of the car and is given by the equation:

F_gravity = m * g,

where m is the mass of the car and g is the acceleration due to gravity (approximated as 9.8 m/s²).

Next, we need to consider the centrifugal force, which is the force pushing the car towards the outside of the circular path. At the top of the hill, the centrifugal force is equal to the gravitational force acting downwards:

F_centrifugal = F_gravity.

The centrifugal force is also given by the equation:

F_centrifugal = (m * v²) / r,

where v is the velocity of the car and r is the radius of the circular path (in this case, the radius of the hill).

Setting the two equations equal to each other and solving for v, we can find the maximum velocity:

(m * v²) / r = m * g.

By rearranging the equation, we get:

v² = r * g.

Taking the square root of both sides, we find:

v = √(r * g).

Substituting the given values, the maximum speed a car can travel over the top of a hill with a radius of 22 m before flying off is:

v = √(22 * 9.8) ≈ 14.93 m/s.

Therefore, the maximum speed is approximately 14.93 m/s.