A Bobcat and a Falling Rock: An Unexpected Encounter

How fast was the rock going just as it bounced off his head?

This problem can be solved using the principles of conservation of energy, assuming that all the potential energy of the rock at the height of 76 meters is converted to kinetic energy just before it strikes the bobcat's head. We can use the equation:

Given data:

Mass of the bobcat = 45 kg

Distance between the bobcat and the rock = 76 meters

Acceleration due to gravity (g) = 9.8 m/s^2

Answer:

The rock was going with a velocity of about 30.1 m/s just before it bounced off the bobcat's head.

When the rock dislodged and started falling from a height of 76 meters above the bobcat, it gained kinetic energy as it converted its potential energy from the height into motion towards the bobcat. The velocity of the rock just before it struck the bobcat's head can be calculated using the conservation of energy equation mentioned earlier:

Potential energy = Kinetic energy

mgh = (1/2)mv^2

By substituting the values of mass of the rock, acceleration due to gravity, and height, we can solve for the velocity (v) of the rock:

m = unknown (mass of the rock)

g = 9.8 m/s^2

h = 76 m

v = unknown (velocity of the rock)

By rearranging the equation, we get:

v^2 = 2gh

v = sqrt(2gh)

v = sqrt(2 * 9.8 * 76)

v = sqrt(1496.8)

v ≈ 30.1 m/s

Therefore, the rock was traveling at a velocity of approximately 30.1 m/s just before it struck the bobcat's head.