A rock of mass is thrown from the edge of a cliff of height with an initial velocity at an angle with the horizontal, as shown in the figure. point is the highest point in the rock’s trajectory, and point is level with the initial position of the rock. all frictional forces are considered to be negligible. which of the following could correctly describe the total energy of the rock-earth system at points and?

Question: A rock of mass is thrown from the edge of a cliff of height with an initial velocity at an angle with the horizontal, as shown in the figure. point is the highest point in the rock’s trajectory, and point is level with the initial position of the rock. all frictional forces are considered to be negligible. which of the following could correctly describe the total energy of the rock-earth system at points and?

We will explore the concept of energy conservation in the context of a simple projectile motion problem. Suppose a rock of mass m is thrown from the edge of a cliff of height h with an initial velocity v at an angle θ with the horizontal, as shown in the figure. Point A is the highest point in the rock’s trajectory, and point B is level with the initial position of the rock. All frictional forces are considered to be negligible. Which of the following could correctly describe the total energy of the rock-earth system at points A and B?

A) The total energy is constant and equal to the kinetic energy at point A.

B) The total energy is constant and equal to the potential energy at point B.

C) The total energy is constant and equal to the sum of the kinetic and potential energies at point B.

D) The total energy is constant and equal to the sum of the kinetic and potential energies at point A.

The correct answer is C. The total energy of the rock-earth system is conserved, meaning that it does not change throughout the motion. This is because there are no external forces doing work on the system. Therefore, the total energy at any point is equal to the initial total energy, which is given by:

E = mv^2/2 + mgh

At point A, the rock has zero vertical velocity, so its kinetic energy is only due to its horizontal component:

EA = mv^2 cos^2 θ/2 + mgh

At point B, the rock has zero height, so its potential energy is zero:

EB = mv^2/2

Since EA = EB, we can conclude that:

mv^2 cos^2 θ/2 + mgh = mv^2/2

Simplifying, we get:

gh = mv^2(1 - cos^2 θ)/2

This equation shows that the total energy depends only on the initial speed and angle of launch, and not on the mass or height of the cliff. Therefore, option C correctly describes the total energy of the system at both points.

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