A uniform rope of length 4.9 m and mass 2.0 kg hangs from a ceiling…

Step 1. Tension and speed. The rope has $\mu = 2.0\ \text{kg} / 4.9\ \text{m} = 0.408\ \text{kg/m}$. At a height $y$ above the block, $T = (M + \mu y)g$, so

Step 2. Speeds (a, b). At $y = 0$: $v = \sqrt{9.8\ \text{m/s}^2 \times 4.9\ \text{m}} = 6.93\ \text{m/s}$. At $y = 4.9\ \text{m}$: $v = \sqrt{9.8\ \text{m/s}^2 \times 9.8\ \text{m}} = 9.80\ \text{m/s}$.

Step 3. Time (c).

Step 4. Half the time (d). The same integral up to a height $y$ is $\dfrac{2}{\sqrt g}(\sqrt{a + y} - \sqrt a)$. Setting it to half the total gives $\sqrt{a + y} = \tfrac12(\sqrt a + \sqrt{a + L})$, so $y = 2.24\ \text{m}$. Less than half the rope: the pulse is slow near the bottom.

Step 5. Wavelengths (e). The frequency stays the same, so $\lambda_{\text{top}}/\lambda_{\text{bottom}} = v_{\text{top}}/v_{\text{bottom}} = \sqrt{(a + L)/a} = 1.41$.