Over a flat plane, how much further can I throw a Major League baseball than I can an M67 grenade?

  • Major League baseball: 148 grams

  • M67 frag grenade: 453 grams

For a baseline of how hard a human can throw a ball (by "I" in this question, understand that I mean anyone), lets assume that this random quote on the internet is accurate:

we know that at the moment of release a ball has about 1/6th horsepower of energy


Someone just told me he could throw a baseball 80 yds, so he assumed he could throw a grenade roughly 50 yds. I'm skeptical. It would be interesting to see a graph on the correlation between an object's weight and the distance it can be thrown based on a given force.

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    $\begingroup$ The quote "assumed to be accurate" is nonsense in terms of units... $\endgroup$ – DJohnM Oct 29 '14 at 5:47
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    $\begingroup$ Horsepower is a unit of power, not of energy. $\endgroup$ – ACuriousMind Oct 29 '14 at 11:17
  • $\begingroup$ Oh I understand. @User58220 It's the only measurement I could find. $\endgroup$ – Viziionary Oct 29 '14 at 15:31

BTW a horsepower, even though an archaic unit retrieved from archaeological digs, is still recognisable to physicists as a unit of power (rate of working) rather than energy.

To answer this question, you need an accurate model of your throwing apparatus: i.e. you need a functional characterisation of the force that can be imparted by the hand as a function of the arm's speed at different positions throughout the throwing movement.

You also need a drag co-efficient for the different things thrown.

The arm with muscles itself has considerable mass and moment of inertia, so I am highly skeptical of the accuracy of the assumption that the ball leaves the hand with a constant energy. The moment of inertia and mass of small balls will be overwhelmed by that of the arm, so the more appropriate model for small balls would be something like a constant velocity release. Moreover, you can greatly increase the amount of work done on the ball by the arm and muscles with a woomera, suggesting the importance of the force versus time versus position dependence in setting the launch conditions.

But let's make some plausible assumptions and see what you get. Neglecting air resistance (probably good for the grenade, not so good for the baseball), if the ball is launched with horizontal and vertical velocity components $v_x(0),\,v_y(0)$ respectively, the time of flight is given, by solving $\dot{v}_y = -g$ by $T = 2\,v_y(0)/g$. In this time, the ball's horizontal flight distance is $2\,v_x(0)\,v_y(0)/g$. If the magnitude of the velocity is $v$, then this horizontal distance is $2\,v^2\,\cos\theta\,\sin\theta$, where $\theta$ is the launch angle (relative to horizontal), maximised at $\theta = \pi/4$. So our horizontal flight distance is:


What this equation says is that the flight distance is proportional to $T/m$, where $T$ is the kinetic energy. So the throwing distance is inversely proportional to the mass.

If we assume the ball is released with constant energy as your folkloric comment seems to assume, then this means that the grenade will fly about a third ($148/453$) of the distance of the baseball.

If we assume that the arm can impart a fixed impulse, then $m\,v$ is the given value at launch. (1) says that the flight distance is $p^2/(m^2\,g)$, i.e. inversely proportional to the square of the mass. Your grenade will fly only a ninth of the distance (you would better run fast in the opposite direction when you throw it!). This doesn't sound right either.

Given the fact that the arm is a constant, significant moment of inertia and mass, I suspect that the relationship is more like the first (inversely proportional to the mass), but indeed that the dependence is somewhat weaker: possibly something like $H\propto \frac{1}{m+m_0}$, where $m_0$ is some fixed mass term. So I suspect the right answer is more like the grenade can be thrown roughly half as far as the baseball. Add to this the fact the grenade is less affected by air resistance, and, surprisingly (I thought like you), I find your friend's assertion highly plausible.

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    $\begingroup$ If nothing else, I would think that another competing model would be "constant torque applied to 'arm plus projectile' system" and your constant of proportionality would be something like $m_{a} + 3 m_{p}$ (assuming that the arm is a uniform rod) $\endgroup$ – Jerry Schirmer Oct 29 '14 at 16:38
  • $\begingroup$ @JerrySchirmer Are you talking about this? $\endgroup$ – WetSavannaAnimal Oct 30 '14 at 3:10
  • $\begingroup$ Nowhere in your answer do you factor in motivation. I'm pretty sure I would throw a grenade pretty darn far away from me... $\endgroup$ – Oscar Bravo Oct 20 '15 at 9:37

An olympic shot put shot is 7260 grams and has been thrown 25 yards. 50 yards for a 453 gram grenade is reasonable.

The military pentathlon has a 600 gram grenade event that requires precision throwing to 35 meters.

Hartmut Nienbar threw a 600 gram grenade 80.3 meters in 1983.

A relatively light-weight object will insignificantly slow the release speed of your arm. So for an adult the release speed of a wiffle ball, tennis ball, baseball, softball will be similar.


Using the method of similarity (see Landau or Arnold), let's assume we can apply the same force to both objects. Therefore, \begin{equation} m\frac{d^2x}{dt^2}=m'\frac{d^2x'}{dt^2}. \end{equation} By reescaling $x'\rightarrow\frac{m}{m'}x$, we get that one can throw the ball $\frac{453}{148}=3,06$ longer than the granade.

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    $\begingroup$ So maybe we could throw a grain of sand to the Moon! $\endgroup$ – DavePhD Oct 29 '14 at 16:37
  • $\begingroup$ With no drag you definitely can :) $\endgroup$ – Diracology Oct 29 '14 at 16:45
  • $\begingroup$ @DavePhD No, I don't believe we can (Diracology). Unless you have a highly extensible arm, your sand would need, even without an atmosphere, to reach the escape speed from the Earth of $11.2{\rm km\, s^{-2}}$. That thought gives me a sore shoulder just thinking about it, not to mention the physiotherapy and rehabilitation bills to heal an arm whose hand had accelerated to $11.2{\rm km\, s^{-2}}$ and back over a few metres. I rather think they'd just cut it off given the damage that would be done! $\endgroup$ – WetSavannaAnimal Oct 30 '14 at 3:17
  • $\begingroup$ I totally agree with you Rod Vance, what I meant is just that if you have no drag force, the similarity method I used for granade and ball would work also for the grain of sand =) $\endgroup$ – Diracology Oct 30 '14 at 11:02
  • $\begingroup$ @WetSavannaAnimalakaRodVance We all agree that that we can't through a grain of sand to the moon, my comment was made to show that the answer is incorrect, that distance thrown is not inversely proportional to mass. $\endgroup$ – DavePhD Oct 30 '14 at 14:18

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