How high will a 1.85-kg rock go from the point of release if thrown straight up by someone who does 80.0 J of work on it? Neglect air resistance.

The particle's potential energy can be described as the energy that is held by the particle because of its position relative to a reference.

The gravitational potential energy is directly proportional to the particle's weight.

Given data:

The mass of the rock is \(m = 1.85\;{\rm{kg}}\).

The work done by gravity is \({W_{\rm{G}}} = 80.0\;{\rm{J}}\).

Work is done on the rock against gravity in lifting the rock from the ground. The rock is moved upward, and the force of gravity acts downward. The rock moves in the vertical direction;so the angle is \(\theta = 180^\circ \).

The expression for the work done is given by:

\( - {W_{\rm{G}}} = mgd\cos \theta \)

Rewrite the above equation.

\(d = \frac{{ - {W_{\rm{G}}}}}{{mg\cos \theta }}\)

Substitute the values in the above expression.

\(\begin{aligned}d &= \frac{{ - \left( {80.0\;{\rm{J}}} \right)}}{{\left( {1.85\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\cos \left( {180^\circ } \right)}}\\d &= 4.41\;{\rm{m}}\end{aligned}\)

Thus, the height of the rock from the point of release is \(4.41\;{\rm{m}}\).