Question

a ladder 25ft long leans against a wall with its foot on level ground 7ft from the base of the wall. If the foot is pulled away from the wall at the rate 2 ft/s express the distance (y) of the top of the ladder above the ground as a function of the time, t seconds in moving

Answer 1

x^2 + y^2 = 625
x = 2t

y^2 = 625 - 4t^2
so,
y = √(625-4t^2)

Answer 2

x^2 + y^2 = 625 (Path. theorem)
x = 7+2t

(7+2t)^2 + y^2 = 625
y^2 = 625 -(7+2t)^2
y=√625 -(7+2t)^2