Assume the force behind my throw to be X. Assume the point of release is 2 meters above ground. Asume the ball is made of shiny steel. I'm not sure if the material matters, I'm just thinking that different materials encounter different air resistance, so I'm assuming the ball is made of steel.

First of all, what would be the optimal angle at which I should throw the ball?

Second of all, what would be the optimal weight of the ball, in order for me to throw it as far as possibile?

I'm interested in finding out the relationship between the angle at which the ball is thrown, the force behind the throw, the ball's mass, the air friction, the point of release, and the distance traveled.

I'm basically interested in the physiscs behind a perfect throw assuming a force X for the throw.

  • 3
    $\begingroup$ A force must be applied for a certain amount of time or a certain distance in order to impart any speed, so that must also be specified in your problem. $\endgroup$
    – FrankH
    Jan 28, 2012 at 16:45
  • $\begingroup$ There are no closed form solutions if there is non-trivial air resistance. A simulation is called for. And it's a decent exercise for a first hand-rolled simulation. $\endgroup$ Jan 28, 2012 at 17:54
  • $\begingroup$ One could postulate that by saying "assuming a force X" the OP really meant "assuming initial energy of the ball as X", as most real-world systems for launching a ballistic projectile will have fixed energy. $\endgroup$
    – dotancohen
    Jan 29, 2012 at 22:12
  • $\begingroup$ Zero. Your speed is therefore c, which is well above galactic escape velocity. Your distance is infinite. $\endgroup$
    – Joshua
    Mar 9, 2016 at 16:23

2 Answers 2


From the 2nd Newton's law one can get the following equations: $$ \left\{ \begin{aligned} \frac{d v_x}{d t} + \frac{F_a(v)}{m v} v_x & = 0 \\ \frac{d v_y}{d t} + \frac{F_a(v)}{m v} v_y & = -g \end{aligned} \right. \qquad (1) $$ where
$v = \sqrt{v_x^2 + v_y^2}$,
$m$ is the mass of the bullet,
$\vec{F}_a$ is the force of the air resistance,
$\vec{g}$ is the free fall acceleration.

If we assume the air resistance force to fit Stoke's law: $$ \vec{F}_a = - 6 \pi \mu R \vec{v} $$ where
$\mu$ is the viscosity,
$R$ is the radius of the ball.

Then system (1) becomes: $$ \left\{ \begin{aligned} \frac{d v_x}{d t} + \frac{6 \pi \mu R}{m} v_x & = 0 \\ \frac{d v_y}{d t} + \frac{6 \pi \mu R}{m} v_y & = -g \end{aligned} \right. \qquad (2) $$ This system can be solved analytically. You can find the solution and investigate it's dependence on the parameters of the ball and the angle.

The coefficient representing air resistance is proportional to $R/m$ rate. If the material is fixed the mass is proportional to $R^3$ and then this coefficient is proportional to $R^{-2}$. This means that the ball should not be too small. Indeed very small particles (dust) loose their velocity very fast.

If the throwing force (initial momentum) is fixed big mass leads to low initial velocity. So the ball should not be too large.

There is an optimal value of the mass and it depends on the angle. So you have to optimize both mass and angle simultaneously.

We can not neglect air resistance because it will lead to zero optimal mass.


Stoke's law does not work for high Reynolds number which can result from high inital velocity and/or large size of the ball. If we use drag equation $$ F_a(v) = \frac{1}{2} \rho v^2 C_D \pi R^2 $$ we will get a system of nonlinear ODEs. That system probably can not be solved analytically.

In this case air resistance will be proportional to $R^{-1}$ and all general conclusions will be the same as for Stoke's law, i.e. both huge and microscopic balls will not fly very far.

  • 2
    $\begingroup$ Stokes' law is not valid for the high Reynold's numbers that will result from throwing a macroscopic object. $\endgroup$
    – Johannes
    Jan 28, 2012 at 19:34
  • 2
    $\begingroup$ Johannes is right, but you could probably substitute it with the drag equation $\vec{F}_a = -\frac{1}{2}C_d\rho Av^2\hat{v}$. $\endgroup$
    – David Z
    Jan 29, 2012 at 5:14

Regarding the question now written as "Assume the force behind my throw to be X."

The problem is not well posed in that you can't use a force to determine an initial velocity -- you also need the time that the force is applied. (One could instead use an initial impulse.) But even if you gave a time that the force is applied, for example, 1/10th of a second, the problem is still unrealistic.

$F=ma$, implies that you can apply an acceleration that approaches infinity as the mass of the ball approaches zero. So if you're looking for an idealistic Newtonian answer, then you should use a zero-mass ball and launch it at infinite speed.

To correct this problem, you need to come up with a limit to how fast the human arm can launch a ball that depends on the mass of the ball, but which does not approach infinity. For example, one way of doing this is to assume that the force is applied not just to the ball but also to some fraction of the mass of the arm. Say the force is applied to a mass equal to the ball plus 1kg.

Under this assumption:
$F = (m+1)a$
and the acceleration of the ball is $a=F/(m+1)$ and the acceleration is still optimized for a zero mass ball but now at least the acceleration is finite. From here, you can work the problem as suggested in other answers.


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.