Final answer:
The car will be airborne for approximately 1.56 seconds, and it will hit the ground approximately 28.08 meters out from the wall. This is a physics problem involving kinematic equations for projectile motion.
Step-by-step explanation:
We will need to use kinematic equations to solve for both the time the car is airborne (a) and the distance it travels horizontally before impact (b). Part a: Time Airborne. Since the car rolls off a cliff horizontally, its initial vertical velocity is 0 m/s. To find the time it's in the air, we'll use the following equation for vertical motion: h = ½ g t^2. where: h is the height of the cliff (12.0 m), g is the acceleration due to gravity (9.81 m/s2), and t is the time in seconds. Plugging in the values, we get t = √(2h/g), t = √(2 * 12.0 m / 9.81 m/s2), t ≈ 1.56 s. Part b: Horizontal Distance. The horizontal distance can be found by multiplying the initial horizontal velocity by the time in the air: d = vh * t. where: d is the distance, vh is the horizontal velocity (18 m/s), and t is the time in the air (1.56 s). So the distance out from the wall is d = 18 m/s * 1.56 s, d ≈ 28.08 m.