Question

A 4 metre ladder is placed against a vertical wall.

Answer 1

Answer:

Original position: base is 1.5 meters away from the wall and the vertical distance from the top end to the ground let it be y and length of the ladder be L.

Step-by-step explanation:

By pythagorean theorem, L^2=y^2+(1.5)^2=y^2+2.25 Eq1.

Final position: base is 2 meters away, and the vertical distance from top end to the ground is y - 0.25 because it falls down the wall 0.25 meters and length of the ladder is also L.

By pythagorean theorem, L^2=(y -0.25)^2+(2)^2=y^2–0.5y+ 0.0625+4=y^2–0.5y+4.0625 Eq 2.

Equating both Eq 1 and Eq 2: y^2+2.25=y^2–0.5y+4.0625

y^2-y^2+0.5y+2.25–4.0625=0

0.5y- 1.8125=0

0.5y=1.8125

y=1.8125/0.5= 3.625

Using Eq 1: L^2=(3.625)^2+2.25=15.390625, L=(15.390625)^1/2= 3.92 meters length of ladder

Using Eq 2: L^2=(3.625)^2–0.5(3.625)+4.0625

L^2=13.140625–0.90625+4.0615=15.390625

L= (15.390625)^1/2= 3.92 meters length of ladder

hope it helps...

correct me if I'm wrong...