Die Another Day

The person originally had $x$ amount of energy to put into the jump. After being shrunk, they'll have $\frac{1}{1000^3}$ of that energy, but they'll have $\frac{1}{1000^3}$ of their original mass (actually all that matters is that the mass and energy reduction factor is the same). This means that, although the initial momentum they can achieve by jumping has decreased, their mass compensates for this decrease and allows them to still travel upwards at the same velocity .

Super Smash Brothers taught me well...

All the work that goes into lifting a person off the ground comes from their muscles which are made out of cells with a given reserve of energy per cell. Given that the man is shrunken down, his muscles will still have the same density of available energy, let's call this density $\rho_E$ .

Thus, the energy available to a man of mass $m$ for jumping will be $E(m)=\rho_E m$ .

With a given kinetic energy $E(m)$ put into his vertical jump, he'll be able to reach the height $h = mg/E(m)$ . Substituting our scaling result for $E$ , we find this height is given by $h = g/\rho_E$ , which is independent of the mass of the man!

Thus, no matter how large or small he is, he'll be able to jump to the same height $g/\rho_E$ .

Hence the maximum height ascended (i.e. 60 cm) will be the same in both cases.

Let's assume that energy available for work is kM, where M is mass and k is some coefficient. If this is converted into gain in gravitational potential energy, then we know kM = gMh, where g is the gravitational constant, and h is height, which works out to k/g. Hence it's independent of mass.

As a matter of fact, both an African lion and an healthy house cat, in jumping at prey above therm, can reach approximately the same altitude (center of gravity to ground). But the lion's reach is longer, giving it the edge in snatching higher prey.

I think my concern with the solutions to this problem are that they assume that the available energy in the muscles is the sole component that decides how high a person can jump. Maybe I'm looking at it the wrong way.

There are two parts to the mechanism of the jump. The first is the available energy in the muscles. The second is the force applied via the legs, which are actually the mechanism of the jump.

Given that the mass will be $1/1000$ of the original, and the energy will be unchanged, I don't see how we can ignore that the legs will be $1/1000$ of the original, thus losing the energy advantage in the loss of mechanical advantage.

If you move a mass M with a lever of length L with some certain energy E, and then use that same energy E to move a mass $M/1000$ with a lever of length $L/1000$ , won't the second displacement be $1/1000$ of the original?

I believe the answer is $.06 cm$ , proportional to the length of the legs, not proportional to the energy in the muscles.

if i'm 190 and jump 60 cm, that means I can jump 3 times less than my height. If i'm reduced to 1000 of my height and still be able to jump 60 cm that would mean I could jump way way more (like 100 times) than my new height, which is practically impossible

It's not clear whether his height is divided by 1,000 or his weight, which initially worried me. However, given that he retains his same density and muscle mass (and assuming we neglect any air resistance effects, which isn't actually stated) it doesn't matter whether his height or his mass is reduced to 1,000th (which will give him different masses and sizes). The equation of force to mass remains the same, so he can still jump 60 cm

$Power = \frac{WD}{time}$

The time taken is somewhat irrelevant, as we assume the same time is taken for muscular contraction at both sizes considering the person is the same level of fitness and hence has the same muscle cell density etc.

Therefore the focus is on work done.

$WD = force * distance$

The force is equal to the weight force of the person. Let $h$ be the jump height, and $m$ be the mass of the person.

$Weight = mg = 9.81m$

$WD = 9.81mh$

Work done is the energy transferred to GPE here. Each muscle cell can transfer a fixed amount of energy in the given time interval, so halving the number of muscle cells will half the work done.

$kWD = 9.81mh$

Getting back to the question, we know that the muscle mass, like the rest of the body, has been reduced by a factor of 1000. Substituting $k$ for $\frac{1}{1000}k$ and $m$ for $\frac{1}{1000}m$ , we see:

$\frac{1}{1000}kWD = \frac{1}{1000}*9.81mh$

Clearly the changes have cancelled each other out, such that the height will be the same always, given that $k$ and $m$ are adjusted to the same degree.

The answer 60 cm would be right (to a crude approximation) if the assumption had been that the amount of ENERGY the muscles could produce was proportional to mass. But the problem said POWER. Assume that 1/1000 means by volume or mass, so that the height of the shrunk person would have been reduced to 1/10. He still must take off at the same speed, but has only 1/10 as much time for his muscles to work before he leaves the ground. So he would have to put out 10 times as much power to reach the same speed in that time. This was a bad problem.

Lets consider 2phases: the leg spread and the jump. Since H=(v2^2-v1^2)/2g, and the speed at the top of the jump is 0, then the jump height is proportional to the square of the speed at which the jump starts. Which brings us to the leg spread phase. Its a constant acceleration from speed 0 to Vmax:

a=Δv/Δt=(vmax-v0)/t=vmax/t

And Vmoy=d/t, also Vmoy=Vmax/2 (constant acceleration)

(1) t=2d/vmax

Then,

(2) a=vmax^2/2d

If the power P is the force F x leg displacement d / time

(3) P=Fd/t

And since F=ma Then,

(4) P=mad/t

Completing eq.4 with eq1 and eq2: P=md(vmax^2/2d)/(2d/vmax) P=mvmax^3/4d

Isolating for vmax,

(5) Vmax=[4Pd/m]^(1/3)

And since :

d2/d1=1/1000

ρ1=ρ2

ρ=m/V ( volume)

V2/V1=10e-9

m2/m1=10e-9

P is proportional to the mass

P2/P1=10e-9

Vmax2/Vmax1=[(P2/P1)(m1/m2)(d2/d1)]^(1/3)=[10e-3]^(1/3)=10e-1

H2/H1=[Vmax2/Vmax1]^2=10e-2

So the jump height is 100 times less high (60cm/100=6mm).

The jump proportional to person height is :

(H2/H1)/(T2/T1)= (10e-2)/(10e-3)=10

So 10 times higher proportional to height (not constant).

When pushing off the ground to jump, the full sized person has acceleration a=F/m, giving him an iniital velocity V. The mass and force of the miniaturized person are both reduced by a factor of 1000, so the ratio F/m doesn't change; the tiny person will have the same initial velocity as the full-sized person, and rise to the same maximum height of 60 cm.

This explains the answer as posed in the problem, but I'm not at all convinced that the assumption about muscle mass and power is correct. But I'm not a physiologist, so please feel free to comment.

$F=m*a$

Let initial * impulse generating acceleration of the person be $ai$ , and the corresponding Force=Fi, *

$Fi=mi*ai$

**When mass is reduced to to 10^-3 Power generated is also reduced to 10^-3 ( Power proportional to Mass)

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Power produced is directly proportional to Force.

Now the impulse generating force of the person can be represented as $F=Fi*10^-3$

Since volume of the person is reduced by 10^-3 the mass should also be reduced by 10^-3 to keep density same.

Then F=mi*10^-3 *a, where a is the impulse generating acceleration .

$Fi*10^-3=mi*10^-3 * a$

$a=Fi/mi$

$a=ai$

impulse acceleration is same which implies the person will acquire same initial Velocity and initial max height !

Let me show you why the answer 60cm is WRONG.

Power P=∆E/∆t=(mgh)/∆t

∆t is the time you need from crouched to stretched legs. Let's suppose you accelerate at a constant rate from v0=0 to vMAX over a distance d. vMAX is your velocity just before you leave the ground. That means:

∆t=2d/vMAX

(You get that by saying ∆t=d/vAVERAGE. 0,5•vMAX=vAVERAGE because acceleration is const.)

Also because of conservation of energy:

vMAX=(2gh)^0,5

So:

∆t=2d/(2gh)^0,5

So:

P=[mgh(2gh)^0,5]/[2d]

Solving for h we get:

h=[(Pd✓2)/(m•g^1,5)]^(2/3)

In the problem was supposed that P was proportional to the mass of the muscles. As lengths are supposed to scale by 10^-3, Volumes will scale by 10^-9. Because density remains the same the mass of the muscles scale by 10^-9.

So P scales by 10^-9. As m in the equation above is also scaled by 10^-9 this factor gets cancelled.

BUT d, the distance over which you accelerate with the force of your legs, gets also scaled - by the factor 10^-3.

That means, that h changes by the factor (10^-3)^(2/3)=1/100

So the correct answer is 0,6cm