Source: Principles of Physics
Problem
A stone of weight $W$ is launched vertically up from the ground with an initial speed $v_0$. It experiences a constant air drag force $F_d$ throughout its flight.
- What is the maximum height, $h_{max}$, reached?
- What is its speed, $v_f$, just before it hits the ground?
[Q1] $h_{max} = \frac{W v_0^2}{2g(W + F_d)}$ [Q2] $v_f = v_0 \sqrt{\frac{W - F_d}{W + F_d}}$
Let the mass be $m=W/g$. We use the work-energy theorem, $\Delta K = W_{net}$. For the upward trip to maximum height, the work done by the net downward force of gravity and drag is $W_{net} = -(W+F_d)h_{max}$. This equals the change in kinetic energy, $\Delta K = -\frac{1}{2}mv_0^2 = -\frac{W}{2g}v_0^2$. Solving gives $h_{max}$. For the downward trip, the work done by the net downward force of gravity and drag is $W_{net} = (W-F_d)h_{max}$. This equals the change in kinetic energy, $\Delta K = \frac{1}{2}mv_f^2 = \frac{W}{2g}v_f^2$. Substituting the expression for $h_{max}$ and solving for $v_f$ gives the final speed.