What is the calculation for the height of a high jump on the moon?

[Nemo1]

Hi Community,

I have this question and I would like to run thru my theory on working it out.
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So my approach is to work out the scaling of the jump on Earth and calculate it for the moon by the following.

\(\displaystyle \frac{x}{2.35}=\frac{1.6m/s^2}{9.8m/s^2}\) solve for \(\displaystyle x=2.35\left(\frac{1.6m/s^2}{9.8m/s^2}\right)\) Simplified to \(\displaystyle x=\frac{94}{245}\)

The next step would be to divide the original 2.35 metres by \(\displaystyle \frac{94}{245}\) to get the height of the jump on the moon to be equal to 6.35 meters.

This seems low so my other thought has been to take the difference of $9.8$ $-$ $1.6$ $=$ $8.2$ and times this by $2.35$ to get a jump of 19.27 metres.

In each case I have put forward I am not confident that I have the correct answer as I am wondering that mass would play a part in this but as it is the same person doing the same jump it would be irrelevant to the calculation and therefore \(\displaystyle F=ma\) does not fit the equation I need to solve this.

Much appreciated for your time in advance.

Cheers Nemo.

 

[MarkFL]

I think the intended approach here is to use the relations:

Acceleration $a$ is the time rate of change of velocity $v$:

\(\displaystyle a=\d{v}{t}\tag{1}\)

Velocity $v$ is the time rate of change of position $x$:

\(\displaystyle v=\d{x}{t}\tag{2}\)

You would assume that the high jumper would have the same upward initial velocity $v_0$ in both cases. So, begin with (1) and derive a general formula for $v$, where $t$ is a variable, and $a$ and $v_0$ are parameters. What do you get?

 

[Nemo1]

I think the intended approach here is to use the relations:

Acceleration $a$ is the time rate of change of velocity $v$:

\(\displaystyle a=\d{v}{t}\tag{1}\)

Velocity $v$ is the time rate of change of position $x$:

\(\displaystyle v=\d{x}{t}\tag{2}\)

You would assume that the high jumper would have the same upward initial velocity $v_0$ in both cases. So, begin with (1) and derive a general formula for $v$, where $t$ is a variable, and $a$ and $v_0$ are parameters. What do you get?

Hi Mark,

I have been trying to get my head around this question.

It turns out I am not the only one in my class so the lecturer shared this.

\(\displaystyle \frac{Earth - S=S_0+V_0t-\frac{9.8}{2}t^2}{Moon - S=S_0+V_0t-\frac{1.6}{2}t^2}\) & \(\displaystyle \frac{V=V_0-9.8t}{V=V_0-1.6t}\)

From here I think the way to solve is to find when $V_0$ is on Earth with a jump of $2.35m$ from \(\displaystyle V=V_0-9.8t\) and plug this into the moons equation of \(\displaystyle {S=S_0+V_0t-\frac{1.6}{2}t^2}\)

What this should give me is that when the Velocity on Earth is equal to Zero then the height is at its maximum of $2.35m$ in Earths Gravity.

\(\displaystyle V=2.35-9.8t = -7.45t\)

Putting this into \(\displaystyle {S=S_0-7.45t^2-\frac{1.6}{2}t^2}\)

I am stuck here, I should be able to get the potential energy from the explanation given but cannot see it.Cheers Nemo

 

[MarkFL]

Okay, you know then that here on Earth we have:

\(\displaystyle v=-9.8t+v_0\)

\(\displaystyle s=-4.9t^2+v_0t\)

We take $s_0=0$ for simplicity, by setting the ground level as zero height. We need to find $v_0$, because this will apply alos to jumping on the moon. As you stated, when the jumper reaches their maximum height, we know $v=0$, and so this implies:

\(\displaystyle t=\frac{v_0}{9.8}\)

And so if we put that into the second equation, and use \(\displaystyle s=2.35\), we obtain:

\(\displaystyle 2.35=-4.9\left(\frac{v_0}{9.8}\right)^2+v_0\left(\frac{v_0}{9.8}\right)\)

If we multiply though by $9.8^2$, we have:

\(\displaystyle 225.694=-4.9v_0^2+9.8v_0^2=4.9v_0^2\)

Dividing through by $4.9$, we now have:

\(\displaystyle v_0^2=46.06\)

Hence:

\(\displaystyle v_0=\sqrt{46.06}=\frac{7}{10}\sqrt{94}\)

Now we are off to the moon...here we know:

\(\displaystyle v=-1.6t+v_0\)

\(\displaystyle s=-0.8t^2+v_0t\)

So, what you want to do now, is plug in for $v_0$, and use the first equation to determine how long it take for the jumper to reach the maximum height (i.e. when $v=0$), and then use your value of $v_0$ and $t$ in the second equation to determine $s$, which will be the maximum height of the jumper on the moon. :)

 

[Nemo1]

Hi Mark,

I have been working thru this question and I am not confident I am getting it.

When we have the initial velocity on Earth $V_0$ and we know the man will jump at that velocity for $t seconds$ and travel $2.35metres$.

Are we trying to find the average velocity for the jump from $0$ to $2.35metres$ and then we can use that average velocity to be able to plug into the lower gravity on the moon and see the man jump higher.

I think my biggest issue is seeing the link between velocity, time and acceleration and how it applies to the different gravities on the Earth and moon.

Thanks for all your teaching so far, I really appreciate your efforts!

Cheers Nemo.

 

[MarkFL]

We assume the jumper has the same initial velocity on the moon as he/she did on Earth, which we found. And so we then need to find how long it takes the jumper to reach the maximum height, which we do by setting the velocity to zero:

\(\displaystyle -1.6t+v_0=0\)

Solve for $t$:

\(\displaystyle t=\frac{v_0}{1.6}\)

Now, we plug this into the function expressing the jumper's height as a function of time:

\(\displaystyle s\left(\frac{v_0}{1.6}\right)=-0.8\left(\frac{v_0}{1.6}\right)^2+v_0\left(\frac{v_0}{1.6}\right)=\frac{v_0^2}{3.2}=\frac{46.06}{3.2}=14.39375\)

As we would expect, this is equal to the jumper's leap height on Earth times the ratio of the Earth's gravitational acceleration divided by the Moon's gravitational acceleration:

\(\displaystyle s_{\text{M}}=s_{\text{E}}\cdot\frac{g_{\text{E}}}{g_{\text{M}}}=\frac{2.35\cdot9.8}{1.6}=14.39375\)

As practice, I would suggest trying to work the problem using parameters rather than numbers. Let the jumper's leap height on Earth be $s_{\text{E}}$, the gravitational acceleration of the Earth be $g_{\text{E}}$, and the gravitational acceleration of the Moon $g_{\text{M}}$. See if you can demonstrate that:

\(\displaystyle s_{\text{M}}=s_{\text{E}}\cdot\frac{g_{\text{E}}}{g_{\text{M}}}\)

 

[MarkFL]

Rather than approaching this problem from a dynamics perspective, we could use energy considerations.

Let's consider the jumper here on Earth, at the moment of launch...they have no gravitational potential energy, but they have kinetic energy:

\(\displaystyle E_i=\frac{1}{2}mv_0^2\)

Next, we consider the jumper the moment they are at the apex of their trajectory...they have gravitational potential energy, but no kinetic energy:

\(\displaystyle E_f=mg_{\text{E}}s_{\text{E}}\)

Using conservation of energy, we conclude that the initial energy is equal in magnitude to the final energy:

\(\displaystyle \frac{1}{2}mv_0^2=mg_{\text{E}}s_{\text{E}}\)

And this reduces to:

\(\displaystyle v_0^2=2g_{\text{E}}s_{\text{E}}\)

Now, if we do the same thing on the moon, we would get:

\(\displaystyle s_{\text{M}}=\frac{v_0^2}{2g_{\text{M}}}\)

Subbing in for $v_0^2$, there results:

\(\displaystyle s_{\text{M}}=\frac{2g_{\text{E}}s_{\text{E}}}{2g_{\text{M}}}=s_{\text{E}}\cdot\frac{g_{\text{E}}}{g_{\text{M}}}\)