Recently I was throwing stones(don't ask me why) when I noticed that there seems to be an optimum weight of stone so that it travels the farthest. If I generate the same amount of force each time(and assuming all other variables like air resistance, angle of projection etc to be constant) shouldn't a smaller stone be projected with a higher velocity and thus have a higher range. You can try this yourselves. A cricket ball sized object goes farther than a small pebble (consider its size to be similar to a coin)( and also farther than a basketball size object, but that is due to the increased mass). My first thought was that it could be air resistance but shouldn't a larger body experience more air resistance? **(I have doubts whether this is a physics question or more of a biology question.)

  • $\begingroup$ Definitely a physics question. +1 $\endgroup$
    – Steeven
    May 1, 2019 at 17:03
  • $\begingroup$ Similar: physics.stackexchange.com/q/471844 $\endgroup$
    – BowlOfRed
    May 1, 2019 at 17:17
  • $\begingroup$ I have little understanding of how air-resistance is modeled but I think this can certainly be explained via taking into consideration air-resistance. The question looks the same as asking why a feather can't be thrown much farther than a pebble. $\endgroup$
    – user87745
    May 1, 2019 at 17:21
  • 3
    $\begingroup$ @DvijMankad Air resistance might contribute, but isn't critical. The important point is that "If I generate the same amount of force..." isn't really happening. The arm cannot put the same force on a pebble or a coin that it can on a more massive object. $\endgroup$
    – BowlOfRed
    May 1, 2019 at 17:26
  • 1
    $\begingroup$ @DvijMankad The difference between a feather and a pebble isn't the same as the difference between two pebbles when considering air resistance. Density is the important factor which determines the relative wind resistance (more area = more drag, less mass = greater acceleration due to applied force, so you should be able to see how relative density makes the effect of the slowing worse). $\endgroup$
    – JMac
    May 1, 2019 at 18:23

3 Answers 3


It is difficult to apply manually the same force and perform the same work for each throw. There could be some physiological reason for what you observed, it should be tested quantitatively. It is also possible that the angle at which you release the object has unwanted changes depending on the weight.

However, what you experience could be tentatively explained in the following simplified way.

For progressively more heavy objects, probably you reach a weight above which the performed work is approximately constant, say $W$. This is likely what you obtain trying to apply the same force along the distance in which the object is hold by the hand. $W$ is accumulated as kinetic energy $\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the speed at which the object is released by the hand. So, roughly, $v=\sqrt{\frac{2W}{m}}$. The heavier is the object, the smaller is the speed $v$. Increasing even more the weight probably will decrease $W$ (but this is a matter about physiology) and thus $v$ will be even smaller.

In the opposite limit, for light objects, their weight does not matter too much: the speed $v$ is simply the final speed of the hand, when it releases the object. The movement of the arm is not affected too much by the presence of a light object hold in the hand. Thus $v$ reaches a constant value $v_0$.

Now we must discuss the relation between $v$ and the distance at which the object is thrown (at which it falls to the ground), with the assumption that the starting angle is the same. Here the friction of air comes into play. According to Stokes's law (see Wikipedia), the force is proportional to the radius $R$ of the object, $F\propto R$. In turn, the deceleration is $\frac{F}{m}\propto \frac{R}{m}$. If we assume objects with similar density, $m\propto R^3$, thus the deceleration is $\propto \frac{1}{R^2}$. Smaller objects are decelerated more than big objects, at least if their shape is almost spherical and their density is similar.

In conclusion. Progressively more heavy objects tend to make parabolic trajectories not affected by air, but their $v$ decreases with increasing $m$. Progressively more light objects tend to start with the same $v=v_0$, but they are more and more decelerated by air as $m$ decreases. The optimum is in between.

A last comment. The effect of the dimension of the object on air friction can be seen in the formula for the terminal velocity of a sphere, i.e. the velocity approached by the sphere falling for a long time in a fluid. This final velocity is $\propto R^2$. Bigger objects tend to fall faster than smaller object. A human body falling from an airplane reaches a speed of 190 km/h and needs a parachute to survive (see Wikipedia); it reaches the speed in 12 s, corresponding to a fall of roughly 300 m. A small spider falling from the roof of a building reaches its terminal speed of (say) 5 mm/s almost instantaneously, slowly falls along the whole building and reaches the ground without harm. Very small particles, with diameter less than 1 $\mu$m, can stay suspended in the slight turbulence of a gentle breeze permanently.

  • 2
    $\begingroup$ FYI, I don't think Stokes' Law is the proper way to analyze this. I think it would be more appropriate to point them towards the drag equation, since it seems unlikely this is being thrown at such low velocities that the assumptions of Stokes' flow holds. $\endgroup$
    – JMac
    May 1, 2019 at 18:26
  • $\begingroup$ Correct, I agree! However, the trend is the same, and the Stokes law is easier to handle. $\endgroup$ May 2, 2019 at 7:57

Doriano Brogioli has pointed out correctly the important role of air resistance. However, I would like to flesh out one of the details in his answer.

If I generate the same amount of force each time...

In reality, your muscles can apply more force when they are moving more slowly. This is called the "force-velocity relationship". This is often approximated as a linear law,

$$ f = f_0 \left(1- \frac{v}{v_0} \right) \textrm{ for } 0 < v < v_0. $$

When pushing against zero load ($f=0$), your muscles move at velocity $v_0$, and no faster. This is approximately the case for very light objects, so they end up with kinetic energy $T= m v_0^2/2$, as Doriano supposed. Then, when pushing against a heavier load, your muscles cannot move as quickly, and the object gains progressively less kinetic energy.


That is because a smaller stone though experiences less air resistance, it is more affected by the same than the larger stone. A larger stone on the other hand is not affected by gravity as much as the smaller stone is and also has the advantage of inertia of motion which does depend upon mass and size of object (Try standing in front of a truck moving at 110km/hr and a cricket ball moving with the same speed. Please try the latter one first as you can do the first one but only once after which you will not be able to do anything). Also even unknowingly ypu are actually exterting a greater force on the larger stone due to conditional reflexes (though negligible but still mentioned). Thus your answer is given. P.S:- Don't tell anyone I said you to stand in front of a truck. PLEASE. Peace.

  • 2
    $\begingroup$ Could you explain "A larger stone on the other hand is not affected by gravity as much as the smaller stone" $\endgroup$
    – Eagle
    May 1, 2019 at 17:41
  • $\begingroup$ This answer is more relevant if you cannot understand the complexity of the one by Doriano. When I started to write this answer there was no answer to this question but upon my completion I found the said answer. That is the reason I am mentioning this in a comment else it would be mentioned in the answer. $\endgroup$ May 1, 2019 at 17:43
  • $\begingroup$ Okay, Eagle the said line is coincides with that of inertia of motion which affecting the larger stone minimalizes the effect of gravity. $\endgroup$ May 1, 2019 at 17:45
  • $\begingroup$ Also as I was writing my first comment there was no comment but upon my completion I found that a comment was made. That is why I had to answer your doubts with the second comment. $\endgroup$ May 1, 2019 at 17:46
  • 1
    $\begingroup$ @JMac Mass scales as the third power of linear dimension while area scales as the second power. The terminal velocities of larger objects are much higher than smaller ones, holding density and shape constant. Please don’t mislead people with such confidently stated nonsense! $\endgroup$ May 1, 2019 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.