General

[quote=LampShadow]Not entirely, assuming that he jumps at the same angle on both the Moon and Earth, his initial v(x) and v(y) vectors would be the same. By breaking down the problem into x and y vectors, v(x) on the Earth = v(x) on the Moon, so the x directional vectors will cancel out (unless we wanted to assume there's an x directional force difference on the Moon compared to Earth, such as differences in air resistance). Because the x vectors cancel out, they're unnecessary in solving this particular problem. Once the y vectors are determined, we can just use proportions in this case to solve the problem.

Although I will admit that I didn't read that it was horizontal displacement, the method I mentioned is just a more fundamental version of what you explained[/quote]

Ohok, I see where you're coming from.

[quote=Ness]Didn't catch that.

Good call.

So my method would be the only viable option then.[/quote]

Not entirely, assuming that he jumps at the same angle on both the Moon and Earth, his initial v(x) and v(y) vectors would be the same. By breaking down the problem into x and y vectors, v(x) on the Earth = v(x) on the Moon, so the x directional vectors will cancel out (unless we wanted to assume there's an x directional force difference on the Moon compared to Earth, such as differences in air resistance). Because the x vectors cancel out, they're unnecessary in solving this particular problem. Once the y vectors are determined, we can just use proportions in this case to solve the problem.

Although I will admit that I didn't read that it was horizontal displacement, the method I mentioned is just a more fundamental version of what you explained

[quote=cb000]Why are people plugging in horizontal displacement into gravitational potential, which depends on vertical displacement...?[/quote]

Didn't catch that.

Good call.

So my method would be the only viable option then.

Why are people plugging in horizontal displacement into gravitational potential, which depends on vertical displacement...?

[quote=LampShadow]@Ness: Yeah you're definitely overthinking it. Like @prestigechef said, this is just a conservation of energy problem. Just use the equation E = mgh

- The amount of energy the person can expend will be the same whether it's on the Moon or the Earth, so we can assume that to be E.
- His mass will be the same on the Moon/Earth, so we can assume that to be m.
- The gravity is different between the Moon/Earth, so Earth gravity = 9.8 m/s^2 and Moon gravity = 1.7 m/s^2
- The height on the moon is what we're solving for, but on Earth h = 3m

So that leaves two equations:
E = m * 9.8 * 3 => 29.4m
E = m * 1.7 * h => 1.7mh

Because E = E, we can set the equations equal to each other:
29.4m = 1.7mh (the m's will cancel out)
29.4 = 1.7h (simplify)
17.29 = h

So the person can jump 17m on the Moon[/quote]

The problem is dealing with horizontal distance, not vertical.

@Ness: Yeah you're definitely overthinking it. Like @prestigechef said, this is just a conservation of energy problem. Just use the equation E = mgh

- The amount of energy the person can expend will be the same whether it's on the Moon or the Earth, so we can assume that to be E.
- His mass will be the same on the Moon/Earth, so we can assume that to be m.
- The gravity is different between the Moon/Earth, so Earth gravity = 9.8 m/s^2 and Moon gravity = 1.7 m/s^2
- The height on the moon is what we're solving for, but on Earth h = 3m

So that leaves two equations:
E = m * 9.8 * 3 => 29.4m
E = m * 1.7 * h => 1.7mh

Because E = E, we can set the equations equal to each other:
29.4m = 1.7mh (the m's will cancel out)
29.4 = 1.7h (simplify)
17.29 = h

So the person can jump 17m on the Moon

[quote=flamedagger]Do you know how long it took him to cover 3m? Because if not, I don't think there is a way to solve it, at least not with kinematics.[/quote]

Not entirely.

If we assume that the 3 meter distance is the maximum horizontal distance, this would mean that the man began with an initial speed 45 degrees above the horizontal.

Using dx = vot + .5at^2, we can use the initial vertical speed, vosin(45), with the displacement of 0 to obtain time t in terms of vo, which is (vosin(45)) / 4.905

We can substitute this value for t into xd = vocos(45)t for the horizontal distance and speed, resulting in an initial velocity of sqrt(14.715 / (cos(45)*sin(45))), or approximately 5.424942 m/s.

Now that you have the initial velocity of the man, just use your new gravity to find his new air time on the moon, and use that to figure out his new maximum horizontal displacement.

Your result should be 17.311765m.

Do you know how long it took him to cover 3m? Because if not, I don't think there is a way to solve it, at least not with kinematics.

i believe... conservation of energy: E = mgh
m(9.8)(3) = m(1.7)d and you solve for d.

I honestly forgot this crap

he can jump farther than 3m