Not a physics major -- Question about how far we can kick an object

[lucid equations]

TL;DR Summary

Debate question for fun

Currently debating with friends how far varying weighed objects could be punted and the distance we could kick them. Instead of grabbing dumbbells I'd rather use math. Was looking for an equation to use and have been having trouble.

I know the average adult can kick with a force of 1000lbs. I've also seen Distance = speed x time
However I have neither speed or time.

Force = mass × acceleration is more promising since I hypothetically have both force and mass of the object but I don't see how I could get distance this way. I understand this is very vague but is there any equations for getting distance that I've overlooked that could help?

I've seen thrown distance can be calculated with
d = v² * sin(2α) / a but this neither requires weight or force I believe which are the main two parameters for the debate.

Also this is assuming the angle is the same each time and the force is the same each time.

Apologies for such a random question.

 

[jbriggs444]

TL;DR Summary: Debate question for fun

Currently debating with friends how far varying weighed objects could be punted and the distance we could kick them. Instead of grabbing dumbbells I'd rather use math. Was looking for an equation to use and have been having trouble.

I know the average adult can kick with a force of 1000lbs.

"Force" is not an ideal starting point. A strong force applied for a brief duration can result in the same momentum as a weaker force applied for a longer duration. A strong force applied over a short displacement can result in the same kinetic energy as a weaker force applied over a larger displacement.

We have a variety of ways of propelling objects. We can throw with our hands or use variety of levers or whips (atlatl, baseball bat, jai alai, sling, bow). These methods use various approaches to extend the duration or displacement of the propulsion, allowing a weaker force to accomplish a more powerful result.

There is also the concern about projectile mass. One can propel a grain of sand to a high speed with very little effort. But air resistance will prevent it from carrying very far.

If there is a point that I am trying to make, it is that the real world is a complicated place. If we want to use a simple equation, we will have to simplify the physical situation a great deal first. Let me do that for you.


Say, for instance that we have a 1 kilogram boot that can be accelerated by a 10 kilogram force (100 Newtons) over a displacement of 1 meter.

We could use the work energy equation: Work = Force times Displacement to compute that 100 Joules of work will have been done on the 1 kilogram boot.

We could use the formula for kinetic energy $\text{KE} = \frac{1}{2}mv^2$ to compute that the 1 kilogram boot must therefore be moving at about 14 meters per second to have that much kinetic energy.

We could imagine an elastic collision between the boot and a small stationary ball. For an elastic collision, the velocity with which the boot approaches the ball will match the relative velocity with which the ball bounces away from the boot. So the ball should attain a velocity of about 28 meters per second. 14 meters per second from the bounce and another 14 meters per second because we measured that velocity relative to the boot that is still following through at 14 meters per second.

For maximum carry distance (ignoring air resistance and assuming level ground), we will assume that the ball has a 45 degree launch angle. The vertical velocity will be about 20 meters per second (28 times the sine of 45 degrees) and the horizontal velocity will also be about 20 meters per second (28 times the cosine of 45 degrees).

While aloft, the ball will be accelerating downward under gravity at a rate of about 10 meters per second per second. It will take 2 seconds to lose its starting velocity on the way up and 2 more seconds to regain it on the way down. That is a total of 4 seconds in flight.

Dring those 4 seconds, the ball will travel 20 times 4 = 80 meters.

So for this contrived scenario, we can kick the ball 80 meters.

 

[PeroK]

If you kick a large rock, it won't go very far and you'll break your foot. And, if you kick a feather, that won't go very far either. And, you can kick a soccer ball further than a table tennis ball. Hmm?

 

[PeroK]

So for this contrived scenario, we can kick the ball 80 meters.

That's quite a punt!

 

[lucid equations]

"Force" is not an ideal starting point. A strong force applied for a brief duration can result in the same momentum as a weaker force applied for a longer duration. A strong force applied over a short displacement can result in the same kinetic energy as a weaker force applied over a larger displacement.

We have a variety of ways of propelling objects. We can throw with our hands or use variety of levers or whips (atlatl, baseball bat, jai alai, sling, bow). These methods use various approaches to extend the duration or displacement of the propulsion, allowing a weaker force to accomplish a more powerful result.

There is also the concern about projectile mass. One can propel a grain of sand to a high speed with very little effort. But air resistance will prevent it from carrying very far.

If there is a point that I am trying to make, it is that the real world is a complicated place. If we want to use a simple equation, we will have to simplify the physical situation a great deal first. Let me do that for you.


Say, for instance that we have a 1 kilogram boot that can be accelerated by a 10 kilogram force (100 Newtons) over a displacement of 1 meter.

We could use the work energy equation: Work = Force times Displacement to compute that 100 Joules of work will have been done on the 1 kilogram boot.

We could use the formula for kinetic energy $\text{KE} = \frac{1}{2}mv^2$ to compute that the 1 kilogram boot must therefore be moving at about 14 meters per second to have that much kinetic energy.

We could imagine an elastic collision between the boot and a small stationary ball. For an elastic collision, the velocity with which the boot approaches the ball will match the relative velocity with which the ball bounces away from the boot. So the ball should attain a velocity of about 28 meters per second. 14 meters per second from the bounce and another 14 meters per second because we measured that velocity relative to the boot that is still following through at 14 meters per second.

For maximum carry distance (ignoring air resistance and assuming level ground), we will assume that the ball has a 45 degree launch angle. The vertical velocity will be about 20 meters per second (28 times the sine of 45 degrees) and the horizontal velocity will also be about 20 meters per second (28 times the cosine of 45 degrees).

While aloft, the ball will be accelerating downward under gravity at a rate of about 10 meters per second per second. It will take 2 seconds to lose its starting velocity on the way up and 2 more seconds to regain it on the way down. That is a total of 4 seconds in flight.

Dring those 4 seconds, the ball will travel 20 times 4 = 80 meters.

So for this contrived scenario, we can kick the ball 80 meters.

Thank you so much for the help and the explanation. The main debate was about how far someone could kick a 10lb weight and we all had varying answers and it got heated.
Again thank you so much. Have an amazing day I'll go off of this set up - until the next debate.

 

[jbriggs444]

Thank you so much for the help and the explanation. The main debate was about how far someone could kick a 10lb weight and we all had varying answers and it got heated.
Again thank you so much. Have an amazing day I'll go off of this set up - until the next debate.

With a 10 pound (5 kg) weight, some of my assumptions go out the window. That is no longer a "small stationary ball". The projectile is considerably more massive than the boot that we are kicking it with.

The normal mechanic for this would be a "shot put", not a "punt". However, let us continue to use the punt mechanic and see what we get.

We have our 1 kg boot approaching the stationary 5 kg ball at a speed of 14 meters per second.

We begin by calculating the speed of the center of mass. We have 14 kg m/sec of momentum on a mass total of 6 kg. That is about 2.3 meters per second.

The relative velocity with which the boot approaches the center of mass is, accordingly, 11.7 meters per second. More importantly, the relative velocity with which the ball approaches the mass center is 2.3 meters per second. It will depart at this same rate. So the ball will be punted away at a total of 4.6 meters per second (4.7 meters per second if we carry through the calculation with better accuracy).

At our launch angle of 45 degrees, that is 3.3 meters per second vertical and 3.3 meters per second horizontal.

That is about 2/3 of a second in flight. For a horizontal distance covered of about 2.2 meters.

Not very far this time.

Maybe there is a reason that one "punts" a football and "puts" a shot.

 

[lucid equations]

With a 10 pound (5 kg) weight, some of my assumptions go out the window. That is no longer a "small stationary ball". The projectile is considerably more massive than the boot that we are kicking it with.

The normal mechanic for this would be a "shot put", not a "punt". However, let us continue to use the punt mechanic and see what we get.

We have our 1 kg boot approaching the stationary 5 kg ball at a speed of 14 meters per second.

We begin by calculating the speed of the center of mass. We have 14 kg m/sec of momentum on a mass total of 6 kg. That is about 2.3 meters per second.

The relative velocity with which the boot approaches the center of mass is, accordingly, 11.7 meters per second. More importantly, the relative velocity with which the ball approaches the mass center is 2.3 meters per second. It will depart at this same rate. So the ball will be punted away at a total of 4.6 meters per second (4.7 meters per second if we carry through the calculation with better accuracy).

At our launch angle of 45 degrees, that is 3.3 meters per second vertical and 3.3 meters per second horizontal.

That is about 2/3 of a second in flight. For a horizontal distance covered of about 2.2 meters.

Not very far this time.

Maybe there is a reason that one "punts" a football and

With a 10 pound (5 kg) weight, some of my assumptions go out the window. That is no longer a "small stationary ball". The projectile is considerably more massive than the boot that we are kicking it with.

The normal mechanic for this would be a "shot put", not a "punt". However, let us continue to use the punt mechanic and see what we get.

We have our 1 kg boot approaching the stationary 5 kg ball at a speed of 14 meters per second.

We begin by calculating the speed of the center of mass. We have 14 kg m/sec of momentum on a mass total of 6 kg. That is about 2.3 meters per second.

The relative velocity with which the boot approaches the center of mass is, accordingly, 11.7 meters per second. More importantly, the relative velocity with which the ball approaches the mass center is 2.3 meters per second. It will depart at this same rate. So the ball will be punted away at a total of 4.6 meters per second (4.7 meters per second if we carry through the calculation with better accuracy).

At our launch angle of 45 degrees, that is 3.3 meters per second vertical and 3.3 meters per second horizontal.

That is about 2/3 of a second in flight. For a horizontal distance covered of about 2.2 meters.

Not very far this time.

Maybe there is a reason that one "punts" a football and "puts" a shot.

So just for my curiosity this was solved primarily using the equation you said KE = 1/2mv^2 ?

 

[PeroK]

Thank you so much for the help and the explanation. The main debate was about how far someone could kick a 10lb weight and we all had varying answers and it got heated.

The women's shot (as in the shot put athletics field event) weighs just under 9lbs. I wouldn't recommend kicking that. Even if you didn't break your foot, I would guess you'd be lucky to kick it more than a metre.

PS a 5-litre container filled with water is about 5kg (11lbs). I'm not sure you could punt that very far either!

 

[jbriggs444]

So just for my curiosity this was solved primarily using the equation you said KE = 1/2mv^2 ?

The $\text{KE}=\frac{1}{2}mv^2$ equation helps us derive the speed of the boot from the force applied and displacement over which it was accelerated.

A bigger boot would wind up with a lower velocity for the same force over the same distance. But the bigger boot could transfer more velocity to the target shot. You could work out the math on that yourself to see if the trade-off is worthwhile.

There is almost certainly an optimal boot mass for a given force profile and shot mass.

 

[jbriggs444]

The women's shot (as in the shot put athletics field event) weighs just under 9lbs. I wouldn't recommend kicking that. Even if you didn't break your foot, I would guess you'd be lucky to kick it more than a metre.

The collision would likely be inelastic. Kicking balls sometimes leads to irreproducible results.

 

[PeroK]

The collision would likely be inelastic.

The problem is partly that 5kg is too heavy for your leg. With a 5kg weight, there's a sense in which it propels your foot further than your foot propels it. With something that heavy, you'd be better trying to use something like a karate kick. As opposed to a football or soccer kicking style.

 

[kuruman]

A followup to post #3 by @PeroK. I view this as a form of impedance matching. Assuming that the collision is (more or less) elastic and that the ball is initially at rest, you want the effective mass of the kicker to be at rest after the collision so that the ball picks up all of the energy. This happens when the two masses are equal. The mass of the ball is easy to figure out. It's another matter to write the kinetic energy of a kicking leg as $\frac{1}{2}m_{\text{eff.}}V^2$ and make reasonable estimates of $m_{\text{eff.}}$ and $V$.

 

[morrobay]

For this elastic collision where the ball, m2 is at rest: kickers leg ,m1 = 18kg , V1 = 20m/s. So V2 (ball after kick). = 2m1/m1+m2 (V1) = 32m/s. Solving for time in flight: Vy=V2sin45-gt. ( set Vy=0) t= 2.3 sec. So distance (x) = V2cos45)(t) , 32m/s(.707)(2.3s) = 52 meters

 

[PeroK]

For this elastic collision where the ball, m2 is at rest: kickers leg ,m1 = 18kg , V1 = 20m/s. So V2 (ball after kick). = 2m1/m1+m2 (V1) = 32m/s. Solving for time in flight: Vy=V2sin45-gt. ( set Vy=0) t= 2.3 sec. So distance (x) = V2cos45)(t) , 32m/s(.707)(2.3s) = 52 meters

When you can kick a 5-kg mass 52m, make a video and post it! A soccer ball weights about 0.5kg. It's a good effort to kick that 52m. Likewise, an American football is about the same weight and 52m is a good punt or field goal.

 

[PeroK]

To get a truly accurate answer would require an experiment or a sophisticated biomechanical analysis. The simplest analysis would be to assume that for a heavy 5-kg ball, the same impulse could be delivered as for a 0.5 kg ball. That would give a ratio of initial velocities in inverse proportion to the mass (which can't be generally valid, but is a simple estimate in the required range of 0.5 - 5 kg).

The range is proportional to $v^2$ hence inversely proportional to the square of the masses.

If we assume that the range for a 0.5kg ball is about 50m, then the range for a 5kg ball would be 0.5m.

That makes some sense to me. I think you would risk injury kicking something as heavy as that, and be barely able to move it at all.

PS For what it's worth Reddit agrees: do not attempt to kick a heavy ball.

 

[PeroK]

PS I'm at the gym and have taken a look at the medicine balls. You could kick the 1kg ball. But possibly not the 2kg ball. The 4kg and 5kg balls are unkickable.

 

[morrobay]

When you can kick a 5-kg mass 52m, make a video and post it! A soccer ball weights about 0.5kg. It's a good effort to kick that 52m. Likewise, an American football is about the same weight and 52m is a good punt or field goal.

For this elastic collision where the ball, m2 is at rest: kickers leg ,m1 = 18kg , V1 = 20m/s. So V2 (ball after kick). = 2m1/m1+m2 (V1) = 32m/s. Solving for time in flight: Vy=V2sin45-gt. ( set Vy=0) t= 2.3 sec. So distance (x) = V2cos45)(t) , 32m/s(.707)(2.3s) = 52 meters

With the givens above with ball 4.5kg . Distance is 52 m. If the ball is .5kg.then the distance is 77 m. So the 5kg example would be for a mechanical kick.However the .5 kg ball can be kicked by person 77m

 

[PeroK]

With the givens above with ball 4.5kg . Distance is 52 m. If the ball is .5kg.then the distance is 77 m. So this example would be for a mechanical kick

That's clearly nonsensical. If the application of physics results in a nonsensical answer, then the application is wrong. It doesn't matter what your numbers say - it's practically impossible to kick a 5kg mass at all and totally impossible to kick it 52m.

 

[morrobay]

That's clearly nonsensical. If the application of physics results in a nonsensical answer, then the application is wrong. It doesn't matter what your numbers say - it's practically impossible to kick a 5kg mass at all and totally impossible to kick it 52m.

Yes biologically. But like I qualified above this example would be for a mechanical device kick.

 

[PeroK]

Yes biologically. But like I qualified above this example would be for a mechanical device kick.

It still looks like nonsense. A humanoid robot (of similar size and mass) would have similar constraints to a human. Firing something out of a cannon would be different - but we are talking about kicking, not using a cannon or a slingshot or using a giant piece of machinery or anything else.

 

[morrobay]

Well @jbriggs444 got 80 meters and I got 77 meters for the .5kg soccer ball. So in the ballpark values according to this link.

 

[PeroK]

Well @jbriggs444 got 80 meters and I got 77 meters for the .5kg soccer ball. So in the ballpark values according to this link.View attachment 350238

No one is disputing how far a 0.5kg ball can be kicked. The question is how far a goalkeeper could kick a 5kg ball.

If you go down to your local gym and see what a 5kg ball feels like, you'll see the problem - and the absurdity that a professional goalkeeper could kick it 50m.

 

[jbriggs444]

Well @jbriggs444 got 80 meters

In case it was not clear, I do not make the claim that my result was biologicaly reasonable. As I understand it, you make no such claim either.

Mine was a back-of-the-envelope estimate based on toy assumptions. I am pleased that it came within a factor of two of a reasonable result.