Hey Kwanga, good to see you again. To find the work done by air resistance, we can use the Work-Energy Theorem, which states that the net work done on an object equals its change in kinetic energy. The net work includes work done by gravity and work done by air resistance. Step 1: Identify the given values. Mass of the ball, $m = 3 \text{ kg}$ Height, $h = 20 \text{ m}$ Initial velocity (dropped), $v_i = 0 \text{ m/s}$ Final velocity, $v_f = 17 \text{ m/s}$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$ Step 2: Calculate the initial kinetic energy ($KE_i$). $$KE_i = \frac{1}{2}mv_i^2$$ $$KE_i = \frac{1}{2}(3 \text{ kg})(0 \text{ m/s})^2$$ $$KE_i = 0 \text{ J}$$ Step 3: Calculate the final kinetic energy ($KE_f$). $$KE_f = \frac{1}{2}mv_f^2$$ $$KE_f = \frac{1}{2}(3 \text{ kg})(17 \text{ m/s})^2$$ $$KE_f = \frac{1}{2}(3 \text{ kg})(289 \text{ m}^2/\text{s}^2)$$ $$KE_f = 1.5 \times 289 \text{ J}$$ $$KE_f = 433.5 \text{ J}$$ Step 4: Calculate the work done by gravity ($W_g$). $$W_g = mgh$$ $$W_g = (3 \text{ kg})(9.8 \text{ m/s}^2)(20 \text{ m})$$ $$W_g = 588 \text{ J}$$ Step 5: Apply the Work-Energy Theorem. The net work done ($W_{net}$) is the sum of the work done by gravity and the work done by air resistance ($W_{air}$). $$W_{net} = W_g + W_{air}$$ Also, the net work done is equal to the change in kinetic energy: $$W_{net} = \Delta KE = KE_f - KE_i$$ Therefore, $$W_g + W_{air} = KE_f - KE_i$$ $$W_{air} = KE_f - KE_i - W_g$$ Substitute the calculated values: $$W_{air} = 433.5 \text{ J} - 0 \text{ J} - 588 \text{ J}$$ $$W_{air} = -154.5 \text{ J}$$ The work done by air resistance is negative because air resistance opposes the motion of the ball. The work done by air resistance is $\boxed{\text{-154.5 J}}$. Send me the next one 📸