Question

An 82.0 kg skydiver jumps out of a balloon at an altitude of 1 100 m and opens the parachute at an altitude of 160 m.

(a) Assuming that the total retarding force on the diver is constant at 60.0 N with the parachute closed and constant at 3 600 N with the parachute open, what is the speed of the diver when he lands on the ground?

(b) Do you think the sky diver will be injured? Explain.

(c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s?m

(d) How realistic is the assumption that the total retarding force is constant? Explain your answer.

Answer 1

a)

using work energy theorem ,

W net = change in kinetic energy

work done by gravity - work done by retarding force when parachute is closed - work done by retarding force when parachute is open = (1/2)mv^2

(mg)(Hf )- (60)(Hf-Hi) - (3600)(Hff - Hii) = (1/2)(m)V^2

(82*9.8*1100) - 60(1100-160) - 3600(160-0) = (1/2) *82 * v^2

883960 - 56400 - 57600 = 41v^2

544040 = 41 v^2

v = 115.2 m/s

b) yes the sky diver will be injured if it lands with such a large speed

c)

using work energy theorem ,

W net = change in kinetic energy

work done by gravity - work done by retarding force when parachute is closed - work done by retarding force when parachute is open = (1/2)mv^2

(mg)(Hf )- (60)(Hf-Hi) - (3600)(Hff - Hii) = (1/2)(m)V^2

(82*9.8*1100) - 60(1100-h) - 3600(h-0) = (1/2) *82 * (5)^2

883960 - 56400 + 60h - 3600h = 1025

h = 233.5 m

therefore parachute must be opened at a height of 233.5 m to land with a speed of 5 m/s

d) the assumption regarding constant retarding force is not realistic as dragging force generally increases with the increase in the speed of falling object.

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