Acar is parked on a cliff overlooking the ocean on an incline that makes an angle of 22◦ 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 4.4 m/s 2 and travels 54 m to the edge of the cliff. the cliff is 28 m above the ocean. how long is the car in the air? the acceleration of gravity is 9.81 m/s 2 . answer in units of s.

The car is 3.4 s in the air.

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

The vertical position of the car can be obtained by the following equation:

y = y0 + v0 · t · sin α + 1/2 · g · t²

Where:

y = vertical position of car at time t.

y0 = initial vertical position.

v0 = initial velocity.

t = time.

α = launching angle.

g = acceleration of gravity.

The vertical component of the position vector when the car reaches the ground is -28 m (considering the edge of the cliff as the origin of the system of reference) and the initial vertical position is therefore 0 m. The launching angle is 22° below the horizontal (see figure). Then, we only have to find the initial velocity to solve the equation of vertical position for the time of flight.

To find the initial velocity, we have to use two equations: the equation of velocity of the car at the time it reaches the edge of the cliff and the equation of position of the car to find that time:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

If we place the origin of the frame of reference at the point where the car starts rolling, then the initial position is zero. Since the car starts from rest, the initial velocity, v0, is zero. Then, we can find the time it takes the car to travel the 54 m down the incline:

x = x0 + v0 · t + 1/2 · a · t²    (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

54 m = 1/2 · 4.4 m/s² · t²

2 · 54 m / 4.4 m/s² = t²

t = 5.0 s

With this time, we can find the velocity of the car when it reaches the edge of the cliff:

v = v0 + a · t   (v0 = 0)

v = a · t

v = 4.4 m/s² · 5.0 s

v = 22 m/s

Then, the initial velocity of the falling car is 22 m/s. Using the equation of vertical position:

y = y0 + v0 · t · sin α + 1/2 · g · t²   (y0 = 0)

y = v0 · t · sin α + 1/2 · g · t²

-28 m = 22 m/s · t · sin 22° - 1/2 · 9.81 m/s² · t²

0 = 28 m + 22 m/s · t · sin 22° - 4.91 m/s² · t²

Solving the quadratic equation for t using the quadratic formula:

t =3.4 s  (the other values is negative and, thus, discarded).

The car is 3.4 s in the air.

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