# Optimum Angle to Throw a Baseball

cj
My Baseball coach is also my physics instructor. He says that "it is easier to throw a baseball horizontally than it is vertically" -- and that throwing speed varies with elevation angle approximately as:

$$v_0cos(\frac {\theta}{2}) \text{ m/s}$$

$$\text {where }v_0 = \text{the initial velocity when the ball is thrown horizontally, and}$$
$$\theta_0 =\text { the initial elevation angle}$$

$$\text{At what }\theta_0 \text{ do I need to throw the ball to achieve maximum height and, conversely, range?}$$

I have a feeling this is a "classic" physics problem. I can't, however, think of the velocity versus launch angle conditions leading to maximum height or range.

The only facts I can cough-up is that at maximum height, vertical velocity = 0. Also, maximum range occurs when the baseball experiences maximum flight time along with maximum horizontal velocity.

Using these relationships to solve for optimum initial launch angle is stumping me.

airbuzz
The vertical initial speed is $$v_0sin(\theta)$$ and the acceleration is obvously -g. The horizontal speed is always $$v_0cos(\theta)$$ if you don´t consider air drag.
So you may calculate the time for the ball to reach maximum height (v0=0) and to fall back to initial height. Then put the same time in horizontal movement and find the launch range.
In this way you find that the maximum range is when the angle is 45°.
The maximum height is achieved when you have the maximum initial vertical speed, so you must have 90°.
If you want to consider the air drag is a dynamical problem (no more only cinematic) and you have to solve it numerically, it´s a little bit more complicated but it´s feasable.
Byez

cj
airbuzz said:
The vertical initial speed is $$v_0sin(\theta)$$ and the acceleration is obvously -g. The horizontal speed is always $$v_0cos(\theta)$$ if you don´t consider air drag.
So you may calculate the time for the ball to reach maximum height (v0=0) and to fall back to initial height. Then put the same time in horizontal movement and find the launch range.
In this way you find that the maximum range is when the angle is 45°.
The maximum height is achieved when you have the maximum initial vertical speed, so you must have 90°.
If you want to consider the air drag is a dynamical problem (no more only cinematic) and you have to solve it numerically, it´s a little bit more complicated but it´s feasable.
Byez

The above certainly holds for basic 2D kinematics.
Having a angle-dependant initial velocity makes the
problem quite a bit more complex -- I THINK.

The expression I saw for the answer is quite complex.
I need to, however, understand the derivation.

Thanks

cj

Mentor
cj said:
The above certainly holds for basic 2D kinematics.
Having a angle-dependant initial velocity makes the
problem quite a bit more complex -- I THINK.
It's still just 2D.

The expression I saw for the answer is quite complex.
I need to, however, understand the derivation.
It's not that bad. Given the initial velocity, find the total time the ball is in the air. (Analyze the vertical component of motion.) Then use that to find the horizontal distance traveled in that time. Once you get that expression, which will depend on the angle, use a little calculus to find it's maximum value.

airbuzz
It´s not so hard.
You have that $$v_y(0) = v_0sin\theta$$ and that $$v_y(t)=v_y(0)-gt$$. Then ,to findthe time to rise to maximum height you mus put $$v_y$$ to zero. The time to rise and fall back will be the double so it will be $$t=\frac{2v_osin\theta}{g}$$. Now you put this time to find the range, multiplying it for the horizontal speed. So you find $$x=V_xt=v_0cos\theta\frac{2v_osin\theta}{g}$$. To find the maximum of x you must just derive x on theta. If your $$v_0$$ is theta-dependant it is not a proble, you must just implement it in the x function. So you will have

$$x=\left(v_0cos\frac{\theta}{2}\right)^2\cdot cos\theta\frac{2sin\theta}{g}$$

now you must integrate this on theta and the job is done!

Mentor
Define "easier." For an infielder throwing to first, you always want the ball to get where you are throwing it as fast as possible. So the optimum angle is whatever angle you need to throw it to get it to your target when throwing it as hard as you can.

If you're an outfielder trying to make a play at home, the problem is much more difficult: with a higher angle, you can get the ball home yourself (after a couple of bounces), but the ball may get there faster if you hit your cutoff man.

Knowing how hard you throw the ball and making a few assumptions about the cutoff man's turn-around time and the slowing of the ball as it bounces, you could calculate whether or not to hit the cutoff man or try for home yourself.

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airbuzz
Problem is coming really interesting...
It become a trajectory optimization problem (deviated mind of an aerospace engineer). So, you have the initial speed depending on something and you want to reach a point in the less time than possible. I use to do this for orbit transfers, but it becomes a numerical boundary conditions problem...

Mentor
airbuzz said:
Problem is coming really interesting...
It become a trajectory optimization problem (deviated mind of an aerospace engineer).
Yeah, we engineers can suck the fun out of damn near anything! airbuzz
The cutoff man should become then a path constraint, like:if you pass there you will have this, but I don´t know if it´s a good thing...
Yeah, with some optimization algorithm it should be feasable... Ehi cj, try to post this problem to your physics professor, maybe he likes this., or maybe he will consider you a fool... :rofl:

Homework Helper
cj said:
The above certainly holds for basic 2D kinematics.
Having a angle-dependant initial velocity makes the
problem quite a bit more complex -- I THINK.

The expression I saw for the answer is quite complex.
I need to, however, understand the derivation.

Thanks

cj

The inital velocity of a ball thrown at angle $$\phi$$ is going to be:
$$\vec{v}_i=<v_0 \cos \frac{\phi}{2} \cos \frac{\phi}, v_0 \cos \frac{\phi}{2} \sin{\phi}>$$
(This follows from the inital velocity, and simple trig.)

Then, to find the peak initial y velocity take the inital y velocity:
$$v_y=v_0 \cos \frac{\phi}{2} \sin{\phi}$$
and find all local minima by solving for the derviative being equal to zero:
$$\frac{dv_y}{d\phi}=v_0(\cos \frac{\phi}{2} \cos{\phi} - \sin\frac{\phi}{2} \sin \phi)=v_0\cos \frac{3\phi}{2}$$
so
$$0=\cos \frac{3\phi}{2} \Rightarrow \cos^{-1} 0 = \frac{3\phi}{2}$$
so
$$\frac{\pi}{2}=\frac{3\phi}{2} \Rightarrow \phi=\frac{\pi}{3}$$
or
$$\frac{3\pi}{2}=\frac{3\phi}{2} \Rightarrow \phi=\pi$$
but the second is obviously incorrect, so the optimal inital launch angle for height is
$$\frac{\pi}{3}$$

Now, the distance that a ballistic projectile travels is
$$\frac{v_i^2}{g} \sin(2\phi)$$
plugging in the values we have for the initial velocity
$$d=\frac{v_0^2}{g} (\cos(\frac{\phi}{2}))^2 \sin(2\phi)$$
Once again, you need to find the maximum - so take the derivative etc. I'm too lazy to go through do the derivative though.

cj
airbuzz said:
The cutoff man should become then a path constraint, like:if you pass there you will have this, but I don´t know if it´s a good thing...
Yeah, with some optimization algorithm it should be feasable... Ehi cj, try to post this problem to your physics professor, maybe he likes this., or maybe he will consider you a fool... :rofl:

His basis was a combination of physics and sports performace lab stuff (which he co-runs at the University of Michigan).

As a former center fielder myself, I know first-hand that
throwing horizontally is clearly "easier" than vertically.

At one point there seemed to be pretty widespread
consideration of what angle an outfielder should toss
a ball to throw out a runner heading for home plate.