a) Let y: vertical distance from the ground
By special right triangle: y = √3/2·z
dy/dt = √3/2·dz/dt ; with dz/dt = 6
dy/dt = 3√3 m/sec
b) Let f: distance to foot of wall
By Law of Cosines: f2 = z2 + 32 - 6zcos60 ; f2 = z2 - 3z + 9
2f·df/dt = 2z·dz/dt - 3·dz/dt ; when z = 3 then f = 3 also.
6df/dt = 18
df/dt = 3 m/sec