Question

ignoring the effects of air, rank the paths according to initial speed.
path 1
path 2
path 3

Explanation:

Step1: Analyze vertical motion

For projectile motion, maximum height $h = \frac{v_{y0}^2}{2g}$, so $v_{y0} = \sqrt{2gh}$. All paths have same $h$, so equal vertical initial velocity $v_{y0}$.

Step2: Analyze horizontal motion

Range $R = v_{x0}t$. Time of flight $t = \frac{2v_{y0}}{g}$, so $t$ is same for all paths. Larger $R$ means larger $v_{x0}$.

Step3: Total initial speed

Total initial speed $v_0 = \sqrt{v_{x0}^2 + v_{y0}^2}$. Since $v_{y0}$ is equal, larger $v_{x0}$ (from larger range) gives larger $v_0$. Path 3 has largest $R$, Path 2 middle, Path 1 smallest.

Answer:

Path 1: 3 (lowest initial speed)
Path 2: 2 (middle initial speed)
Path 3: 1 (highest initial speed)
Ranking (from highest to lowest initial speed): Path 3 > Path 2 > Path 1