We can solve this problem using the conservation of energy principle, which states that the total mechanical energy (potential energy + kinetic energy) of a system remains constant if no external work is done on the system.
At the top of the hill, the car has only potential energy, given by:
PE = mgh
where m is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill.
PE = (1000 kg) x (9.81 m/s^2) x (200 m) = 1,962,000 J
At the base of the hill, the car has only kinetic energy, given by:
KE = (1/2)mv^2
where v is the velocity of the car.
Using the conservation of energy principle, we can equate the potential energy at the top of the hill to the kinetic energy at the base of the hill:
PE = KE
mgh = (1/2)mv^2
Canceling the mass on both sides, we get:
gh = (1/2)v^2
Solving for v, we get:
v = sqrt(2gh)
Substituting the given values, we get:
v = sqrt(2 x 9.81 m/s^2 x 200 m) = 44.3 m/s
Therefore, the velocity of the car at the base of the hill is approximately 44.3 m/s.