SOLUTION: A car is parked on a cliff overlooking the ocean
on an incline that makes an angle of 18.0◦
below the horizontal. The negligent driver
leaves the car in neutral, and the e
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on an incline that makes an angle of 18.0◦
below the horizontal. The negligent driver
leaves the car in neutral, and the e
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Question 214250: A car is parked on a cliff overlooking the ocean
on an incline that makes an angle of 18.0◦
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 2.00 m/s2 and travels 43.0 m to the edge of
the cliff. The cliff is 40.0 m above the ocean.
The acceleration of gravity is 9.81 m/s2 .
a) How long is the car in the air? Answer
in units of s.
b) What is the car’s position relative to the
base of the cliff when the car lands in the
ocean? Answer in units of m.
whats steps would u take to solve this problem Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A car is parked on a cliff overlooking the ocean
on an incline that makes an angle of 18.0◦
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 2.00 m/s2 and travels 43.0 m to the edge of
the cliff. The cliff is 40.0 m above the ocean.
The acceleration of gravity is 9.81 m/s2 .
a) How long is the car in the air? Answer
in units of s.
b) What is the car’s position relative to the
base of the cliff when the car lands in the
ocean? Answer in units of m.
-----------------------------
s = (at^2)/2
43 = (2*t^2)/2
t = sqrt(43) seconds to go the 43 meters
v = at = 2sqrt(43) m/sec when the car leaves the incline and is airborne.
v =~ 13.114877 m/sec
------------
The vertical component of v is v*sin(18) = -4.05272 m/sec (neg since is down)
The horizontal part of v is v*cos(18) = 12.473 m/sec
---------------------
h(t) = -9.81t^2 - 4.05272t + 40
h(0) = 40 (when it leaves the ground)
Find t at h = 0 (when it hits the water)
-9.81t^2 - 4.05272t + 40 = 0
Quadratic equation (in our case ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=1586.0245393984 is greater than zero. That means that there are two solutions: .
Quadratic expression can be factored:
Again, the answer is: -2.23637327194893, 1.8232519671579.
Here's your graph:
t =~ 1.823252 seconds
----------------------
The horizontal part of v is v*cos(18) = 12.473 m/sec
12.473 m/sec * 1.823252 seconds =~ 22.741 meters from the base of the cliff
