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The curious question of the leaping kangaroo

  • Thread starter zoogies
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A friend sent me this problem over IM. I've never seen anything quite like this before and even after she explained it (albeit a rushed explanation), I still have little idea what parts of the problem mean, much less how the answer was arrived at.
The problem is this:
A 40 kg kangaroo exerts a constant force on the ground in the first 60 cm of her jump, and rises 2 m higher. When she carries a baby kangaroo in her pouch, she can rise only 1.8 m higher. What is the mass for the baby kangaroo?
So, first of all, what does it mean to exert a force in the first 60cm of a jump? I can understand easily exerting a force for six seconds, or six milliseconds. But first 60cm? Does the kangaroo somehow exert a constant force on the ground even after she is 60cm in the air? Does the velocity time graph during this period of constant force then look something like so:

Code:
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in that at first the net force (force of her jump minus mg) carries her continually upward until the area beneath the curve is 0.6m, and thereafter it becomes the graph of an object in free fall?

(update)

Uhm, apparently, the 0.6m motion is from when the kangaroo is crouching down to when the legs are fully straightened, so the kangaroo has not left the ground in any way. Also apparently, there are initial and final velocities associated with that, although I can't imagine how, if the kangaroo doesn't...actually...move...does it? x.x
 
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Answers and Replies

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ehild
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zoogies said:
Uhm, apparently, the 0.6m motion is from when the kangaroo is crouching down to when the legs are fully straightened, so the kangaroo has not left the ground in any way. Also apparently, there are initial and final velocities associated with that, although I can't imagine how, if the kangaroo doesn't...actually...move...does it? x.x
The feet stay on the ground while the center of mass moved upward by 0.6 m, and it will rise 2 m high without the baby....
ehild
 

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