5.A car approaches the top of a hill that is shaped like a vertical circle with a radius of 55.5 m. What is the fastest speed that the car can go over the hill without losing contact with the ground?

To find the fastest speed that the car can go over the hill without losing contact with the ground, we need to consider the forces acting on the car at the top of the hill.

At the top of the hill, the car experiences two forces: the gravitational force (mg) acting downward, and the normal force (N) exerted by the ground, acting perpendicular to the surface.

To prevent the car from losing contact with the ground, the normal force must provide enough centripetal force to keep the car moving in a circular path. At the top of the hill, the net force acting towards the center of the circle is given by F_net = N - mg.

The centripetal force required to keep the car moving in a circle of radius 55.5 m is given by F_cen = (mv^2)/r, where m is the mass of the car, v is its speed, and r is the radius of the circle.

Setting the net force (F_net) equal to the centripetal force (F_cen) gives us:

N - mg = (mv^2)/r

Now, we can solve for the speed (v). First, multiply both sides of the equation by r:

Nr - mrg = mv^2

Next, divide both sides of the equation by m:

v^2 = (Nr - mrg)/m

Finally, take the square root of both sides to find the speed (v):

v = sqrt((Nr - mrg)/m)

Given the radius of the hill (r = 55.5 m), we need to know the mass of the car (m) and the acceleration due to gravity (g) to calculate the fastest speed.