A body of mass, m is projected vertically upward with a velocity, y. At what position will the potential energy be maximum?

Analyze the question: A body is thrown straight up with speed "y" (the question uses y for the initial velocity). Potential energy is greatest at the highest point, where the body momentarily stops. We need that height.

Concept: At maximum height, all the kinetic energy has changed into potential energy (conservation of energy). Equivalently, we can use the equation of motion v² = u² − 2gh.

Solution:

Using energy conservation:

• Initial kinetic energy = ½my²
• Potential energy at maximum height h = mgh

Set them equal:

½my² = mgh

Cancel m:

h = y²/(2g)

Check with the equation of motion: at the top, final velocity v = 0, so 0 = y² − 2gh, which gives h = y²/(2g).

The potential energy is maximum at a height of y²/(2g).

Answer: Option C

Tip: "Maximum height" in vertical projection is always u²/(2g). Remember that at the top the velocity is zero, so v² = u² − 2gh quickly gives the height. Also notice that the mass cancels out, so the height does not depend on m. Option A (y/g) is the time to reach the top, which is a common distractor.