The curious question of the leaping kangaroo

[zoogies]

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:

/

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

 

[ehild]

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

 

[quicknote]

I can help clarify the concept of force and its role in the kangaroo's jump. When the kangaroo is crouching down, it is exerting a constant force on the ground in order to push off and jump upwards. This force is what allows the kangaroo to overcome the force of gravity and rise 2m in height. The 60cm refers to the distance over which this constant force is exerted.

When the kangaroo carries a baby in her pouch, the extra mass affects the amount of force she can exert on the ground, resulting in a lower jump height of 1.8m. By using the principles of force and motion, we can calculate the mass of the baby kangaroo by setting up equations and solving for the unknown variable.

It is important to note that the velocity-time graph will look different for the kangaroo's jump with and without the baby. The initial and final velocities will also vary in each scenario.

I hope this helps clarify the problem and the role of force in the kangaroo's jump. If you have any further questions, please let me know. Science is all about asking questions and seeking answers, so don't hesitate to reach out for more information.