Question

A rope 26 meters long passes through a pulley h meters above the ground on a flagpole. On one end of the rope is a weight that is w meters above the ground and at the other end is a person holding the rope x meters from the flagpole; see picture. flagpole pulley rope h weight person ground We can relate the quantities x,h,w via the equation: 26=h-w+ x2+h2 (a) When the pulley is 9 meters above the ground and the weight is 4 meters above the ground, the x-value is (b) Assume the pulley is moving up the pole at a constant rate of 2/3 meters/sec and the person moves away from the pole at a constant rate of 3/4 meters/sec. When the pulley is 9 meters above the ground and the weight is 4 meters above the ground, how quickly is the weight rising into the air?

Answer 1

26= h-wt J x² 2 2 for h=9 w=4 26= 9-4 + x²+ 81 ✓ x²+81 2) x²78 441 360 6J 10 & miface dw + ako sh.dh dh de ot A du Differentiating given equation dh di dt at 21 2²7 2² dw dh x drilt th dhielt dt dt + x²th + 6 JT () + (3) (%) dw 21 et 42 alt It 4ot9 to