This has been posted before but I need someone to elaborate please.
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 19.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 3.06 m/s2 for a distance of 50.0 m to the edge of the cliff, which is 50.0 m above the ocean. Find the following.
(a) The car's position relative to the base of the cliff when the car lands in the ocean.
(b) The length of time the car is in the air.
What's given:
Cliff angle with the horizontal, θ = -19°
Distance on the cliff, d = 50m
Height of cliff, h = -50m
(0 being on top of the cliff)
acceleration before leaving cliff, a = 3.06 m/s²
acceleration due to gravity, g = -9.8 m/s²
Car was stationary at the beginning, so
initial velocity, u = 0
First calculate the speed of the car, v, when it left the cliff using
v²-u² = 2ad
At the point of leaving the cliff,
horizontal velocity, vh = vcos(θ)
vertical velocity, vv = vsin(θ)
(note the sign of θ)
We now can calculate the time in the air, t using
h = vv*t + (1/2)gt²
The horizontal distance from the cliff, S can be calculated by the product of the initial horizontal velocity, vh by the time in the air.
Post if you need explanations or answer check.
I'm not sure how to find a)
To solve this problem, we can break it down into two parts:
(a) Finding the car's position relative to the base of the cliff when it lands in the ocean.
(b) Finding the length of time the car is in the air.
(a) Finding the car's position relative to the base of the cliff when it lands in the ocean:
To find the car's position relative to the base of the cliff, we need to find the horizontal and vertical components of its motion separately.
The horizontal component of the car's motion remains constant since there is no force acting on it in the horizontal direction. Therefore, the horizontal distance covered by the car is given by:
distance_horizontal = horizontal_velocity * time
To find the horizontal velocity, we can use the formula:
horizontal_velocity = initial_velocity * cos(angle)
Given that the car starts from rest, the initial velocity is 0. Therefore:
horizontal_velocity = 0 * cos(19°) = 0
So the horizontal distance covered by the car is 0.
The vertical component of the car's motion can be determined using the kinematic equation:
distance_vertical = initial_height + (initial_velocity * time) + (0.5 * acceleration * time^2)
Given that the initial height is 50.0 m and the acceleration is -3.06 m/s^2 (negative because it's moving downward), we can plug in these values:
distance_vertical = 50.0 + (0 * time) + (0.5 * -3.06 * time^2)
Now, we need to find the time it takes for the car to reach the edge of the cliff. To find this, we can use the formula for time:
time = sqrt((2 * distance) / acceleration)
Given that the distance is 50.0 m, we can calculate the time:
time = sqrt((2 * 50.0) / -3.06) ≈ 4.92 seconds
Now, we can substitute the calculated time back into the equation for distance_vertical:
distance_vertical = 50.0 + (0 * 4.92) + (0.5 * -3.06 * (4.92)^2)
Using this equation, we can find the vertical distance covered by the car relative to the base of the cliff when it lands in the ocean.
(b) Finding the length of time the car is in the air:
To find the length of time the car is in the air, we need to determine at what time the car reaches the edge of the cliff.
We have already calculated the time it takes for the car to reach the edge of the cliff, which is approximately 4.92 seconds. Therefore, the car is in the air for the same duration, 4.92 seconds.
By following these steps, you can find the car's position relative to the base of the cliff when it lands in the ocean and the length of time the car is in the air.