Posted by **Pia** on Monday, July 8, 2013 at 4:17pm.

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

- Maths -
**Steve**, Monday, July 8, 2013 at 4:59pm
x^2 + y^2 = 625

x = 2t

y^2 = 625 - 4t^2

so,

y = √(625-4t^2)

- Maths -
**M. Mubashir Bhatti**, Saturday, February 8, 2014 at 4:27am
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

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