The top of a 20-foot ladder leans against a wall, and the bottom of the ladder slides away from the wall at a rate of 5 ft/sec. find the velocity of the top of the ladder at time t=1 sec if the bottom of the ladder is 7 feet away from the wall at time t=0

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.

that would be because of both gravity and friction. so     c