How high and far could a person jump in lower gravity?

Question

Specifically, half of Earth's gravity. Let's assume that all other factors are equal to Earth.

At a glance, it seems like the answer might simply be "double", but at the same time it seems as though the massive shift of strength to body weight ratio would make things less obvious.

Also, bonus points for how high/far something could be thrown (proportionally) and how gravity plays into the fall speeds of each of these things.

Answer 1

How high is less useful a question than how far.

How Far?

Without even considering the physiology of the jumper, we can immediately consider how the change in gravity would affect horizontal distance for the jumper (or any object). Consider the equation for distance with velocity:

$d = vt$

Where d = distance, v = velocity, and t = time. If we want something to go travel farther, we need to increase velocity and/or time.

Now consider the equation for the fall time of an object starting at a velocity of 0 m/s:

$t = \sqrt{2d/g}$

Where d = distance and g = our gravity. In this scenario, we know that a = 1/2 earth's gravity. With that in mind, we can make the assertion that fall time for an object on the new planet will be approximately 41% ($\sqrt{1/(1/2)}$) longer than the fall time for an object on earth in similar circumstances.

Using information and the earlier equation, we also now know that the travel distance for an object will be about 41% farther than a similar object on earth - minus the effects of drag.

Take Olympic long jumping, for example: the Men's record is held by Bob Beamon 8.9 M, or 29.16 ft. That same person performing on your planet would manage a 12.55 M or 41.12 ft. long jump.

All that being said, there is the issue of physiology in the low-gravity environment, and that will play a role in maximum jump distance. However, once you've figured a jump height and running speed, you can use the previous equations to render it to an approximate answer that assumes a parabolic trajectory.

As for thrown objects, all of that holds true, just with slightly different issues of figuring out throw velocity and angle, then expanding that throw velocity to the vertical speed and horizontal speed. From there, use the following equations for throw time and throw distance:

$t = 2\cdot (V/4.905)$ | V = vertical speed

$d = H\cdot t$ | H = horizontal speed, t = time, solved from above

Answer 2

Assuming you can jump 1 meter high in Earth gravity conditions, your initial velocity would have to be 4.43 m/s.

It turns out that this is a constant, regardless of what the gravitational force is; your legs can't spring any faster just because the gravity is lower, because your muscles and bones can only produce a limited amount of force at a given point in time. (Courtesy of this space exchange question).

To calculate how high we can jump we can use the maximum height formula:

H = (vo^2 * sin ^2 theta) / 2g

In this case, our numbers turn out to be:

H = ((4.43 * 4.43) * (sin ^2 90))/9.8
H = 2.002540816

This means you can jump just over double the height of on earth.

The horizontal range formula is:

range = ((initialvel)^2 * (sin 2theta))/g

Since the best range is when launching at 45 degrees, we plug in numbers to get

range = ((4.43 * 4.43) * (sin (90)) / 4.9
range = 3.05805

Your ratio for height and distance are as you suspected; approximately double the original.

Answer 3

Ignoring air resistance (as this has to do with gravity), it should be as simple as using a vector equation.

Let's say $Sx(t)$ is horizontal position (for how far one could jump), and $Sy(t)$ is vertical position (for how high one could jump). Then it's as simple as this:

1. Determine gravity $(g)$ in distance units per time units squared. (Often meters per second squared, or feet per second squared)
2. Determine initial horizontal velocity $(Vx0)$ and initial vertical velocity $(Vy0)$. This is how fast you're going when you first jump, which is determined by lots of factors in muscles and such.
3. Set $Sx(t) = Vx0 * t + Sx0$, and$Sy(t) = -(g/2) * t^2 + Vy0 * t + Sy0$, where S0 is starting hieght. (If starting on the ground, you can often set S0 to 0)
4. Combine the functions into a vector function, with the points $(Sx(t), Sy(t))$.

This can be done easily in desmos.com/calculator, where you can easily type the x function and y function, and then put them together as an "ordered pair" (vector function) to see how high and far you'd jump.

Jumping at a 45 degree angle results in maximum distance, which simply means that $Vx0 = Vy0$

In short, if you halve gravity, but initial velocity is the same (not guaranteed), you can jump double the distance. But you have to consider how you body works when jumping, and how gravity affects it. Lower gravity would make it more difficult to stay on the surface and push off with the same force, so ir may be less than double. From a strictly numeral basis, though, it can be easily determined with the above equations.

Answer 4

Just to complete the others answers, with a biological approach

Travelling to the new planet: During the travel, the astronauts will lose muscle mass and bone density, even with training (see the astronauts when they come back from the ISS). And to go on an other planet, the voyage will be much more longer, so your travellers will lose much more strength.

Recreating muscles: Nature doesn't do useless thing. The astronauts will do training on the new planet but will never retrieve their Earth strength, cause this is useless on a planet where the gravity is halve. They will have muscle to fight this new gravity not the Earth one. They so will be able to jump as high as on our planet, because more is useless.