Post a New Question


posted by on .

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 - ,

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

    y^2 = 625 - 4t^2
    y = √(625-4t^2)

  • Maths - ,

    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

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question