<span>Bones. The most important organ of the skeletal system is the bones.
Ligaments and Joints. Another important component, i.e. the ligaments are made of fibrous collagen tissue that attaches one bone to another bone.
<span>
Cartilage.</span></span>
The distance it traveled and the time that it took to travel that distance
Well first of all, you must realize that it depends on how the jumpers are distributed on the earth's surface. If,say, one billion of them are in the eastern hemisphere and the other billion are in the western one, then the sum of all of their momenta could easily be zero, and have no effect at all on the planet. I'm pretty sure what you must have in mind is to consider the Earth to be a block, with a flat upper surface, and all the people jump in the same direction.
average mass per person = 60 kg.
jump velocity = 7 m/s straight up and away from the block, all in the same direction
one person's worth of momentum = (m) (v) = 420 kg.m/s
sum of two billion of them = 8.4 x 10¹¹ kg-m/s all in the same direction
Earth's "recoil" momentum = 8.4 x 10¹¹ in the opposite direction = (m) (v)
Divide each side by 'm' : v = (momentum) / (mass) =
The Earth's "recoil" velocity is (8.4 x 10¹¹) / (5.98 x 10²⁴) =
1.405 x 10⁻¹³ m/s =
<em> 0.00000000014 millimeter per second
</em>I have no intuitive feeling for this kind of thing, so can't judge whether
the answer is reasonable. But my math and physics felt OK on the
way to the solution, so that's my answer and I'm sticking to it.
To solve the problem it is necessary to apply conservation of the moment and conservation of energy.
By conservation of the moment we know that

Where
M=Heavier mass
V = Velocity of heavier mass
m = lighter mass
v = velocity of lighter mass
That equation in function of the velocity of heavier mass is

Also we have that 
On the other hand we have from law of conservation of energy that

Where,
W_f = Work made by friction
KE = Kinetic Force
Applying this equation in heavier object.






Here we can apply the law of conservation of energy for light mass, then

Replacing the value of 

Deleting constants,

