A car is parked near a cliff overlooking the
ocean on an incline that makes an angle of
21.7◦ 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 20 m/s when it reaches the edge of the cliff. The cliff is 19.5 m above the ocean. The acceleration of gravity is 9.8 m/s. How far is the car from the base of the cliff
when the car hits the ocean?
To solve this problem, we can break it down into several steps:
Step 1: Find the time it takes for the car to reach the edge of the cliff.
Using the equation of linear motion:
v = u + at
where v is the final velocity (20 m/s), u is the initial velocity (0 m/s), a is the acceleration (due to gravity, -9.8 m/s²), and t is the time we need to find.
Let's rearrange the equation to solve for t:
t = (v - u) / a
t = (20 - 0) / -9.8
t ≈ 2.04 seconds
Step 2: Find the horizontal distance traveled by the car.
Using the equation of motion for uniformly accelerated motion:
s = ut + (1/2)at²
where s is the distance, u is the initial velocity (0 m/s), a is the acceleration (-9.8 m/s²), and t is the time (2.04 seconds).
Let's substitute the values:
s = 0 * 2.04 + (1/2)(-9.8)(2.04)²
s ≈ -20.04 meters
(Note: The negative sign indicates that the distance is in the opposite direction of the inclined plane.)
Step 3: Find the vertical distance traveled by the car.
The vertical distance is given as 19.5 meters.
Step 4: Find the total distance from the base of the cliff.
Using the Pythagorean theorem:
distance² = horizontal distance² + vertical distance²
Let's substitute the values:
distance² = (-20.04)² + 19.5²
distance² ≈ 802.4016 + 380.25
distance² ≈ 1182.6516
distance ≈ √1182.6516
distance ≈ 34.38 meters
Therefore, the car is approximately 34.38 meters from the base of the cliff when it hits the ocean.
To solve this problem, we can use the equations of motion for objects in free fall.
First, we need to find the time it takes for the car to reach the edge of the cliff. We can use the equation:
v = u + at
Where:
v = final velocity (20 m/s)
u = initial velocity (0 m/s, since the car starts from rest)
a = acceleration (due to gravity, -9.8 m/s^2)
t = time
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Plugging in the values, we get:
t = (20 - 0) / -9.8
t ≈ -2.04 seconds
Since time cannot be negative, we discard the negative sign. Therefore, the car takes approximately 2.04 seconds to reach the edge of the cliff.
Next, we can use the equation of motion to find the horizontal displacement (distance) of the car:
s = ut + 0.5at^2
Where:
s = horizontal displacement (distance)
u = initial horizontal velocity (0 m/s, since there is no acceleration in the horizontal direction)
a = horizontal acceleration (0 m/s^2, since there is no acceleration in the horizontal direction)
t = time (2.04 seconds)
Since a = 0, the equation simplifies to:
s = ut
Plugging in the values, we get:
s = 0 * 2.04
s = 0 meters
Therefore, the car is 0 meters from the base of the cliff when it hits the ocean.
figure the time to go to the bottom
Hf=hi+vi*t+1/2 g t^2
0=19.5-20sin21.7 t -1/2 g t^2
solve for t.
how far? distance= 20cos21.7 * t