Final answer:
The car was traveling at approximately 39.59 m/s when it left the cliff.
Step-by-step explanation:
To find the speed of the car when it left the cliff, we need to use the principle of conservation of energy.
At the top of the cliff, the car has potential energy due to its height above the ground. When it falls, this potential energy is converted into kinetic energy.
The formula for potential energy is PE = mgh, where m is the mass of the car (which we can assume to be constant), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the cliff (80 meters).
The formula for kinetic energy is KE = 1/2 mv^2, where m is the mass of the car again, and v is its velocity.
Since energy is conserved, the potential energy at the top of the cliff is equal to the kinetic energy when the car lands.
So we can set mgh = 1/2 mv^2 and solve for v:
mgh = 1/2 mv^2
Canceling out the mass:
gh = 1/2 v^2
Solving for v:
v = √(2gh)
Plugging in the given values, we get v = √(2 * 9.8 * 80) ≈ 39.59 m/s.
So the car was traveling at approximately 39.59 m/s when it left the cliff.