WAEC - Physics (2024 - No. 24)

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

\(\frac{y}{g}\)

\(\frac{2y^2}{g}\)

\(\frac{y^2}{2g}\)

\(\frac{y^2}{4g}\)

At the maximum height, all the kinetic energy will have been converted into potential energy:

mgh = \(\frac{1}{2}\)mv\(^2\)

v = y

gh = \(\frac{1}{2}\)y\(^2\)

h = \(\frac{y^2}{2g}\)

Thus, the potential energy will be maximum at the height given by: \(\frac{y^2}{2g}\)