Your friends car is parked on a cliff overlooking the ocean on an incline that makes an angle of 17.0° below the horizontal. The brakes fail, and the car rolls from rest down the incline for a distance of 29.0m to the edge of the cliff, which is 55.0m above the ocean, and, unfortunately, continues over the edge and lands in the ocean.
a) Find the car's position relative to the base of the cliff when the car lands in the ocean.
b) Find the length of time the car is in the air.
h = 29*sin17 = 8.48 m. = ht. of the incline.
V1^2 = Vo + 2g*h = 0 + 19.6*8.48 = 166.2, V1 = 12.9 m/s[17o] = Velocity at bottom of the incline.
X1 = 12.9*Cos17 = 12.3 m/s.
Y1 = 12.9*sin17 = 3.77 m/s.
Y2^2 = Y1^2 + 2g*h = 3.77^2 + 19.6*55 = 1092.2, Y2 = 33 m/s. = Final velocity.
Y2 = Y1 + g*t = 33, 3.77 + 9.8t = 33, 9.8t = 29.23, t = 2.98 s.
a. Dx = X1*t = 12.3 * 2.98 = 36.7 m.
b. T = 2.98 s (Calculated above).
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To solve this problem, we can break it down into two parts: the horizontal displacement and the vertical displacement of the car.
a) Finding the horizontal displacement:
The horizontal displacement can be found using the distance and the angle.
Given:
Distance (d) = 29.0 m
Angle (θ) = 17.0° below the horizontal
To find the horizontal displacement (x), we can use the equation:
x = d * cos(θ)
Substituting the given values:
x = 29.0 m * cos(17.0°)
Calculating the horizontal displacement, we get:
x ≈ 27.94 m
Therefore, the car's horizontal position relative to the base of the cliff when it lands in the ocean is approximately 27.94 meters.
b) Finding the time the car is in the air:
The time the car is in the air can be found using the vertical displacement and the acceleration due to gravity.
Given:
Vertical displacement (y) = 55.0 m
To find the time (t), we can use the equation for vertical displacement:
y = (1/2) * g * t^2
Where g is the acceleration due to gravity.
Rearranging the equation to solve for time t, we get:
t = sqrt( (2 * y) / g )
Substituting the given values:
t = sqrt( (2 * 55.0 m) / 9.8 m/s^2 )
Calculating the time, we get:
t ≈ 3.18 s
Therefore, the car is in the air for approximately 3.18 seconds.
a) To find the car's position relative to the base of the cliff when it lands in the ocean, we can break down the horizontal and vertical components of the car's motion.
First, let's determine the horizontal distance the car travels before landing in the ocean. The angle of 17.0° below the horizontal makes the vertical component of motion irrelevant for this part, as it only affects the car's time in the air. The horizontal distance traveled can be found using the equation:
x = d * cos(θ)
where x is the horizontal distance, d is the total distance traveled down the incline (29.0m), and θ is the angle of incline (17.0°).
Plugging in the values, we have:
x = 29.0m * cos(17.0°)
Using a calculator, we find that x ≈ 27.62m.
This means the car will land approximately 27.62m horizontally away from the base of the cliff.
Now, let's determine the vertical distance the car falls before landing in the ocean. The vertical distance can be found using the equation:
y = d * sin(θ)
where y is the vertical distance (height), d is the total distance traveled down the incline (29.0m), and θ is the angle of incline (17.0°).
Plugging in the values, we have:
y = 29.0m * sin(17.0°)
Using a calculator, we find that y ≈ 8.06m.
This means the car will fall approximately 8.06m vertically from the edge of the cliff.
So, the car's position relative to the base of the cliff when it lands in the ocean is approximately 27.62m horizontally away and 8.06m vertically below the base of the cliff.
b) To find the length of time the car is in the air, we can use the vertical motion equations. The only force acting on the car in the vertical direction is gravity, which causes it to accelerate downward.
The equation that relates the vertical distance (y), initial vertical velocity (v₀), time (t), and acceleration due to gravity (g) is:
y = v₀t + (1/2)gt²
Since the car starts from rest (v₀ = 0), we can simplify the equation to:
y = (1/2)gt²
We have already found that y is approximately 8.06m. The acceleration due to gravity is approximately 9.8 m/s².
Plugging in these values, we have:
8.06m = (1/2)(9.8 m/s²)t²
Simplifying the equation further, we get:
t² = (2 * 8.06m) / (9.8 m/s²)
t² ≈ 1.65
Taking the square root of both sides of the equation, we find:
t ≈ 1.28 seconds
Therefore, the length of time the car is in the air is approximately 1.28 seconds.