As a rock of mass 3.0 kg drops from the edge of a 60. meter high cliff, it experiences air resistance, whose average strength during the fall is 22. N. At what speed will the rock hit the ground?

Answer

1.5 m/s

13 m/s

18 m/s

26 m/s

34 m/s

1.5 m/s

To find the speed at which the rock hits the ground, we can use the concept of "energy conservation."

The total mechanical energy of the system (rock + Earth) remains constant throughout the fall because there are no external forces doing work on the system (ignoring air resistance for now).

The initial potential energy of the rock at the top of the cliff is converted into kinetic energy as it falls toward the ground.

The potential energy (PE) of an object at a height h can be calculated using the formula:
PE = mgh

Where:
m = mass of the rock (3.0 kg)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the cliff (60 meters)

So, the initial potential energy of the rock is:
PE = (3.0 kg)(9.8 m/s^2)(60 m) = 1764 J

The final kinetic energy (KE) of the rock just before it hits the ground is equal to the initial potential energy. Therefore,
KE = 1764 J

The kinetic energy of an object can be calculated using the formula:
KE = (1/2)mv^2

Where:
m = mass of the rock (3.0 kg)
v = speed of the rock

Substituting the known values into the equation, we get:
(1/2)(3.0 kg)(v^2) = 1764 J

Simplifying the equation, we have:
v^2 = (2 * 1764 J) / 3.0 kg
v^2 = 3528 J / 3.0 kg
v^2 = 1176 m^2/s^2

Taking the square root of both sides, we find:
v ≈ √1176 m^2/s^2
v ≈ 34 m/s

Therefore, the rock will hit the ground with a speed of approximately 34 m/s.

The correct answer is 34 m/s.