An 80.0 kg student running at 3.5 m/s grabs a vertically hanging rope. How high will the student swing (maximum height)?

What the problem is asking

You are finding the maximum vertical rise after the student grabs the rope. At the highest point, the student momentarily stops, so all of the initial running energy has been converted into gravitational potential energy.

Convert kinetic energy into gravitational potential energy

Initial kinetic energy: $$K = \tfrac{1}{2}mv^2$$ Gain in gravitational potential energy at height $h$: $$U = mgh$$ Set them equal (conservation of mechanical energy, neglecting friction and air resistance): $$\tfrac{1}{2}mv^2 = mgh$$

Solve for the height

Cancel $m$ from both sides and solve: $$h = \frac{v^2}{2g}$$ Substitute $v = 3.5\,\text{m/s}$ and $g = 9.8\,\text{m/s}^2$: $$h = \frac{(3.5)^2}{2(9.8)} = \frac{12.25}{19.6} \approx 0.625\,\text{m}$$

Quick reasonableness check

A speed of $3.5\,\text{m/s}$ is moderate, so a rise well under $1\,\text{m}$ makes sense. The mass not affecting the height is also expected because both kinetic and potential energy scale with $m$.