A car is parked near a cliff overlooking the ocean on an incline that makes an angle of 29.1◦ with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has a velocity 6 m/s when it reaches the edge of the cliff. The cliff is 42.3 m above the ocean.

The acceleration of gravity is 9.8 m/s2. How far is the car from the base of the cliff when the car hits the ocean? Answer in units of m.

The initial vertical velocity component is Vyi = -6.0 sin 29.1 = -2.918 m/s

The horizontal velocity component remains
Vx = 6.0 cos 29.1 = 5.243 m/s during the fall.

Solve this quadratic equation for the time T that it takes to fall to the ocean:

Y = 0 = 42.3 -2.918 T -(g/2) T^2
(Take the positive root; there will be two)

Once you have T, get the horizontal distance with
X = Vx * T

Thanks drwls

Awesome job.

Why did the car park near the cliff? Because it wanted a great view! Now let's calculate how far it traveled before hitting the ocean.

Let's use some physics to solve this. We can break down the motion of the car into two components: one parallel to the incline and one perpendicular to it. The perpendicular component won't affect our calculations since it doesn't contribute to the car's horizontal distance traveled.

We can start by finding the acceleration of the car parallel to the incline. We know the final velocity is 6 m/s and the initial velocity is 0 m/s. We also know the acceleration due to gravity is 9.8 m/s². We can use the final velocity equation:

v² = u² + 2as

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

Plugging in the values, we get:

6² = 0² + 2 * a * s

36 = 2as

Next, let's find the component of the acceleration due to gravity acting parallel to the incline. We can use trigonometry to do this. The angle of the incline is 29.1 degrees, so the component of gravity acting parallel to the incline is:

a_parallel = g * sin(angle)

a_parallel = 9.8 m/s² * sin(29.1°)

Now we can equate the two equations for acceleration:

2as = 9.8 m/s² * sin(29.1°)

s = (9.8 m/s² * sin(29.1°)) / (2a)

Now we can plug in the values for the given acceleration due to gravity and the component of acceleration parallel to the incline:

s = (9.8 m/s² * sin(29.1°)) / (2 * 9.8 m/s²)

s = sin(29.1°) / 2

Finally, we can calculate s:

s ≈ 0.25 m

So, the car is approximately 0.25 m from the base of the cliff when it hits the ocean. Just a small splash!

To find the distance the car is from the base of the cliff when it hits the ocean, we can use the equations of motion.

First, let's find the time it takes for the car to reach the edge of the cliff. We know that the car starts from rest, so its initial velocity, u = 0 m/s. The final velocity, v = 6 m/s. The acceleration, a, is due to gravity and is equal to 9.8 m/s^2 (negative because it opposes the motion). We need to find the time, t.

We can use the equation: v = u + at

Substituting the values: 6 = 0 + (-9.8)t

Simplifying the equation: 6 = -9.8t

Dividing both sides by -9.8: t = -0.6122 seconds (approx.)

Note: The negative sign indicates that the car is moving in the opposite direction of the positive y-axis.

Now, let's find the horizontal distance traveled by the car when it reaches the edge of the cliff. We can use the equation for horizontal distance:

Distance = Velocity × Time

Substituting the values: Distance = 6 m/s × t

Distance = 6 m/s × (-0.6122 seconds)

Distance = -3.6732 meters (approx.)

Again, note that the negative sign indicates the car is moving in the opposite direction of the positive x-axis.

Finally, let's find the vertical distance traveled by the car when it hits the ocean. We are given that the cliff is 42.3 meters above the ocean.

Therefore, the total vertical distance traveled by the car is 42.3 meters.

Now, we can use the Pythagorean theorem to find the actual distance from the base of the cliff:

Distance^2 = Horizontal Distance^2 + Vertical Distance^2

Distance^2 = (-3.6732 meters)^2 + (42.3 meters)^2

Distance^2 = 13.5005 + 1790.29

Distance^2 = 1803.79

Taking the square root of both sides, we get:

Distance ≈ 42.49 meters

Thus, the car is approximately 42.49 meters from the base of the cliff when it hits the ocean.