Answer:
vy= -3.75 ft/sec (slides down )
Explanation:
denoting y as the coordinate of the top of the ladder , x as the coordinate of the bottom , L as the length of the ladder , then
x²+y²=L² = constant
deriving by t
2*x*dx/dt + 2*y*dy/dt = 0
since the velocity is vx=dx/dt and vy=dy/dt
x*vx+y*vy=0
then since the velocity of x is constant x=x₀+vx*t
at t=1 sec x=7 ft + 5 ft/sec*1 sec = 12 ft
then x²+y²=L² → y= √(L²-x²) =√[(20 ft)²-(12 ft)²] = 16 ft
after that
x*vx+y*vy=0 → vy= -(x/y)*vx = -(12 ft/16 ft)*5 ft/sec= -3.75 ft/sec
vy= -3.75 ft/sec (slides down since vx is positive, that means that slides away)
Note
we cannot assume that vy is constant ( and do y=y₀+vy*t) when vx is constant , since y= √(L²-x²) = √(L²-(x₀+vx*t)²) , that is not linear with t . Nevertheless, vy could be found deriving y with respect to t , but that would be more complicated than the other way.
