Final answer:
Using physics kinematic equations, the car will be airborne for approximately 1.56 seconds and will travel about 28.08 meters from the wall before hitting the ground.
Step-by-step explanation:
When a car rolls off a 12.0m tall cliff horizontally at 18 m/s, we can use the principles of physics to determine the time it remains airborne and the distance it will travel before it hits the ground.
The time the car is in the air depends only on the vertical motion. Since the car does not have an initial vertical velocity (it rolls off horizontally), we can use the following kinematic equation for free-fall:
where h is the height of the cliff (12.0m), g is the acceleration due to gravity (9.81 m/s²), and t is the time in seconds. Rearranging the equation to solve for t and substituting the known values:
- 12.0m = ½(9.81 m/s²)(t²)
- t² = (2×12.0 m)/(9.81 m/s²)
- t = √[(2×12.0 m)/(9.81 m/s²)]
- t ≈ 1.56 seconds
To find how far out from the wall the car hits the ground, use the horizontal velocity and the time airborne:
- distance = velocity × time
- distance = 18 m/s × 1.56 s
- distance ≈ 28.08 meters
Hence, the car will be airborne for approximately 1.56 seconds and will land about 28.08 meters away from the base of cliff.