To protect his food from hungry bears, a boy scout raises his food pack with a rope that
is thrown over a tree limb at height h above his hands. He walks away from the vertical rope
with constant velocity vboy, holding the free end of the rope in his hands (Fig. P2.71). (a) Show
that the speed v of the food pack is given by x(x2 + h2)–1/2 vboy where x is the distance he has
walked away from the vertical rope. (b) Show that the acceleration a of the food pack is
h2(x2 + h2)–3/2 v2
boy. (c) What values do the acceleration and velocity v have shortly after he
leaves the point under the pack (x = 0)? (d) What values do the pack’s velocity and acceleration
approach as the distance x continues to increase?
All the kinematics equations and Pythagorean Theorem.
The Attempt at a Solution
I have been working at this for awhile now and can't seem to get it. I know that x=vboyt. I know that there is some trig here along the lines, but I can't get how the food box moves. Ok if the boy moves x then l will grow also, but at what rate? However much l grows is how high up the box goes. I can't seem to figure it out.
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