David and Emily are climbing El Capitan, a big cliff wall in Yosemite National Park.  David is on the ground holding the rope attached to a carabiner (a rope “pulley” that is on the wall) above Emily as she climbs.  When Emily stops to rest, David wonders how high she has climbed. The rope is attached to his waist, about 3 feet off the ground, and he has let out 48 feet of rope which goes up to the carabiner and then back down the wall to Emily’s harness.  The rope at David’s waist makes a 55° angle with the ground and he is standing 20 feet away from the base of the wall.

1. Assuming that the rope is taut (pulled tight), approximately how long is the rope between David and the carabiner above Emily?

1. How high up the wall has Emily climbed?  Describe your method.

If D is the end of the rope at David’s end (round his waist) and C is where the rope reaches the carabiner, then DC=48-h where h is the vertical distance CE between Emily (E) and the carabiner.

cos(55)=20/(48-h)=DB/CD because David is 20 ft from the wall and the rope length in total is 48 ft. The legs of a right triangle CDB are 20 (base DB) and CB (wall), with hypotenuse CD and B is the point on the wall opposite D.

(1) Therefore CD=48-h=20/cos(55)=34.8689 ft=34.87ft approx.

(2) h=CE=48-34.8689=13.1311 ft. So Emily is 13.1311 ft from the carabiner. CB=20tan(55)=28.563 ft. So the height of the carabiner is 28.563+3=31.563ft above ground, and Emily’s climbed height is 31.563-13.1311=18.43ft approx.

by Top Rated User (716k points)