The rate at which the kinetic energy ($K$) of a body changes is equivalent to the power ($P$) delivered by the net force acting on it. From the Work-Energy Theorem: $$ \frac{dK}{dt} = \vec{F}_{\text{net}} \cdot \vec{v} $$ The ball is subjected to two forces:

• Gravitational force: $mg$ (downwards)
• Air resistance: $F_{\text{air}} = kv$ (opposing motion)

We must compare the rate of energy change during upward and downward motion.

Upward Motion: The velocity decreases from initial $u$. Both gravity and drag act downwards. $$ |P_{\text{up}}| = (mg + kv)v $$ This is maximum at the moment of projection ($v=u$). $$ |P_{\text{up, max}}| = (1.0 \times 10 + 0.41 \times 4.0) \times 4.0 \approx 46.6 \, \text{W} $$

Downward Motion: The ball accelerates from rest. Gravity acts down, drag acts up. The net force is downwards (accelerating). $$ F_{\text{net}} = mg – kv $$ The rate of change of KE is: $$ P(v) = (mg – kv)v = mgv – kv^2 $$ This value starts at 0 (when $v=0$) and returns to 0 when terminal velocity is reached ($mg = kv$). The maximum must occur in between.

To find the velocity $v$ where $P$ is maximum, we differentiate $P$ with respect to $v$ and set it to zero: $$ \frac{dP}{dv} = \frac{d}{dv}(mgv – kv^2) = mg – 2kv $$ $$ mg – 2kv = 0 $$ $$ v = \frac{mg}{2k} $$

Let’s calculate the maximum power at this velocity to compare with the upward phase: $$ P_{\text{max}} = \left(mg – k \frac{mg}{2k}\right) \frac{mg}{2k} = \frac{mg}{2} \cdot \frac{mg}{2k} = \frac{(mg)^2}{4k} $$ $$ P_{\text{max}} = \frac{(10)^2}{4 \times 0.41} = \frac{100}{1.64} \approx 60.98 \, \text{W} $$

Since $60.98 \, \text{W} > 46.6 \, \text{W}$, the kinetic energy changes most rapidly during the downward journey at the velocity derived above.

Substituting the given values: $$ m = 1.0 \, \text{kg}, \quad g = 10 \, \text{m/s}^2, \quad k = 0.41 \, \text{kg/s} $$ $$ v = \frac{1.0 \times 10}{2 \times 0.41} $$ $$ v = \frac{10}{0.82} $$ $$ v \approx 12.195 \, \text{m/s} $$

Rounding to appropriate significant figures: v ≈ 12.2 m/s.