# Flat throw vs 45 degree throw of a ball

Why does my sports coach tell me that when I'm fielding I should throw the baseball 'flat' to get the maximum distance? I thought from physics that you get the most distance from throwing at a 45 degree angle? What should I be doing to get the maximum distance?

• for baseball, it should be maximum distance in minimum time, but if you throw a Javelin, then it should be at 45 degrees because time doesn't matter there... – Ankit Lamba Aug 13 '15 at 10:06
• A pure physics approach can't say what angle of throw will give you the maximum distance. For a simple catapult that's always launches the object at the same speed you'd be right, 45 degrees will give the longest distance, but the human body is different and much more complicated. You might be able to throw faster horizontally than at 45 degrees. – bdsl Aug 13 '15 at 10:13
• Throwing a ball horizontal at someone that is paying attention makes it almost impossible for that person to not catch it (given that you throw straight at them). Throwing with an arc means you can't throw too hard because the ball will overshoot it's goal. – Kevin Aug 13 '15 at 12:45
• @Kevin You can't throw a ball horizontally. It's in flight a finite time and will drop due to gravity (unless you have an amazing backspin) in that time. – Carl Witthoft Aug 13 '15 at 14:45
• You guys are obviously not ball players. Just because a baseball hits the ground doesn't mean the fielder can no longer catch it. A lower angle takes off less momentum on the bounce (we are playing on ball fields , not some abstract surface ) ... not to mention that the cutoff man couldn't reach it at a 40+ degree angle. Next, can you throw a rock farther than you can skip it? – technosaurus Aug 14 '15 at 6:21

I am not a baseball expert, but the time, seconds, will maybe matter here. You are correct, the maximum distance, theoretically, is when you throw at a 45 degree angle. However, in baseball you might want the maximum distance compared to how long time it will take. You maybe want the ball to get as far as possible, as fast as possible? This might make your coach a bit more correct?

Also, of course, air resistance is in play here too.

• Any reasonable theory about thrown baseballs includes air resistance so, "theoretically" the maximum distance is at an angle that's close to 45 but not exactly there. Please don't use the word "theoretically" to mean "in some impossible situation": the whole point of physics is to come up with theories that accurately model reality. – David Richerby Aug 13 '15 at 13:25
• My objection is to your use of the word "theoretically". If, instead, you'd said something like, "Ignoring air resistance, which doesn't make a huge difference, the maximum distance is when you throw at a 45 degree angle", I'd have been completely happy. – David Richerby Aug 13 '15 at 15:36
• Some comments that were not contributing deleted. Personal remarks are a violation of the overriding "Be nice" policy of all Stack Exchange sites, and remarks and questions explicitly about votes don't actually help to improve the question. – dmckee --- ex-moderator kitten Aug 15 '15 at 16:09

45 degree angle for a projectile gives you the maximum distance in a vacuum, but air resistance, as pointed out, changes that a little. With air resistance slowing the ball, you need to to throw a tick under 45 degrees for maximum distance.

Also

Also, since you throw from above the shoulder, not from the ground, the ball is usually thrown a foot or more above the altitude where it's caught and that shaves another degree or two off the optimal angle for distance, so optimal distance, I'm thinking maybe 40 degrees - as a guess.

I agree with Jakethesnake and I suspect your coach is just teaching what he was told when he played baseball and he's not a student of physics. When you throw straighter, it's a faster throw and that's the desired outcome in what is sometimes referred to as the game of inches. You want to hit the cutoff man or the 3rd baseman, not throw the ball as far as you can.

So, yeah, your coach gets an F in physics, but he knows his baseball.

• Rather a large "tick" :-) forbes.com/sites/stevensalzberg/2013/04/29/… or physics.stackexchange.com/q/52223 – Carl Witthoft Aug 13 '15 at 14:42
• You do realize that you can still catch a throw after it bounces right? The lower the angle the less forward momentum lost on each bounce. – technosaurus Aug 14 '15 at 6:25
• I'm surprised I overlooked that, and yes, you're right. – userLTK Aug 14 '15 at 7:39
• I would suspect that the fact that you are usually throwing overhand also has something to do with it, because when throwing flatter the ball leaves the hand later and therefore you apply force to it longer for greater total impulse. – Jan Hudec Aug 14 '15 at 12:11
• Throwing it longer occurred to me as well, after writing. The elbow only bends back so far, so you likely do get more "oomph" with a straighter throw, more muscular movement, straighter arm extension at the time of release, all that good stuff. – userLTK Aug 14 '15 at 13:03

I am a baseball fan (and a physicst), and your coach is misleading you a little. First, in the absence of air resistance, a 45-degree launch will get the ball there with miminum energy expenditure. But not, as the other answers suggest, minimum time. And time matters. A lot. :-)

Your coach should also be telling you to plan your longer throws such that they bounce once before reaching the destination (typically direct to homeplate), for the same basic reason of minimizing elapsed time. The speed lost due to the bounce causes a time penalty much less than the time it takes to throw a ball in a high-enough arc to reach home on the fly. The tradeoff between direct throw and one bounce depends on the distance and the absolute strength of the outfielder's arm.

It's worth noting that a few third basemen (originally a certain disgraced gambler) will throw a one-bouncer to first base for the same reason.

Why does my sports coach tell me that when I'm fielding I should throw the baseball 'flat' to get the maximum distance? I thought from physics that you get the most distance from throwing at a 45 degree angle?

The second question first, this is true if you are a robot throwing a ball on the Moon (no atmosphere) that releases the ball at the same speed regardless of the release angle. You are a human, not a robot, and you are throwing the ball in the Earth's atmosphere rather than on the airless Moon. You need to account for physiology and realistic conditions. When you do that, you will find that the optimal angle is well under 45 degrees. The initial velocity of a baseball thrown by a human depends on release angle; the highest speed releases are with throws that are close to horizontal. This alone drops the optimal angle to less than 40 degrees. Presumably you are throwing the ball overhand. When you release the ball it will have a good deal of backspin. This gives lift. Accounting for lift and drag drops the optimal angle for maximum distance even more, to 35 degrees or less.

Now for the second question. While your coach's reasoning was incorrect, his advice was spot-on. Your goal is not to throw the ball as far as you can. Your goal is to get the runner out. You will not get the runner who started at first or second base out with a long-distance throw if the distance to home plate is at the very edge of your throwing capabilities. The time aloft will be far too long, and just as importantly, your accuracy will be lousy. You'll be better off throwing to the cutoff man.

Check out this list of eight phenomenal baseball throws made from deep in the outfield at http://www.hardballtimes.com/a-physics-comparison-of-great-throws-from-years-past/. All have three things in common: They were incredibly fast, incredibly accurate, and thrown at a rather low elevation angle, from 5.9 degrees to 14.3 degrees.

This is a physics site, so I'm going to pitch a website dedicated to the physics of baseball, http://baseball.physics.illinois.edu. This is written by an emeritus physics professor at UIUC who also is a big fan of baseball.

• Good answer. Really like the links. Backspin matters quite a lot, I believe. It would be interesting to show how the steeper angle throw takes longer. Without air resistance, time to cover distance $d$ with a throw of velocity $v$ that lands at the same height as it takes off goes as $t = \sqrt{\frac{d\sin\2\alpha}{g}} - so 45 degrees takes 1.7 x longer than 10 degrees... the difference between "safe" and "out". – Floris Aug 13 '15 at 20:11 • Update... see my answer for the correct calculation. There was an extra factor of$\cos\alpha$that I had neglected in my comment. – Floris Aug 13 '15 at 20:51 • +1 for lift / backspin. Didn't see that in any of the higher-voted answers, just air resistance. – Peter Cordes Aug 15 '15 at 5:30 Let's analyze with some simplifications. Ignoring friction for a moment... if you throw a ball at angle$\alpha$with velocity$v$such that it will fly for a distance$d$, and it is caught at the same height as it is released, then we can quickly calculate the velocity and time of flight. Time of flight: $$t = 2\frac{v_y}{g} = \frac{2v\sin\alpha}{g}$$ Distance covered: $$d = v_x t = \frac{2v^2\sin\alpha\cos\alpha}{g}=\frac{v^2}{g}\sin{2\alpha}$$ If you know the distance you want to throw, you can now compute the velocity as a function of angle: $$v = \sqrt{\frac{dg}{\sin{2\alpha}}}$$ and the time taken to cover the distance $$t = \frac{d}{v_x} = \frac{d}{\sqrt{\frac{dg}{2\sin{\alpha}\cos\alpha}}\cos\alpha}\\\bbox[5px, lightblue]{t=\sqrt{\frac{2d\sin\alpha}{g}}}$$ It follows from this that if you have a strong enough arm, the time for your throw will be shorter if you can throw at a lower angle; as David Hammen pointed out, this might even mean throwing with a bounce. For example, let's take a look at the amazing throw by Yoenis Cespedes (link from David Hammen's post), clocked at 101.5 mph with a start angle of 10 degrees and reaching its target in 2.8 seconds. He might have save himself some trouble with a 45 degree lob - but (using all the approximations above) it would have arrived about 2.8 seconds ($\sqrt{\frac{\sin{45}}{\sin{10}}}$) later... when there was just a split second before the Kendrick would have been safe. Now this is a simplification - air resistance, spin, and the difference in height between release and catch will all affect the result. But even the basic physics says that throwing fast and low will get more people out. The "high pitched throw" may be needed to get greater range - but at the top end, outfielders that can launch a 100 mph bullet don't need that... • +1; I refrain myself from discussing because you can't exactly answer this; there are many things responsible like Bernoulli's theorem, drag which is proportional approximately to the velocity but actually it depends on the square too & it is very difficult to find the coefficients as they not only depend on the dimension of the ball but also how viscous the air is at the required moment; of course it doesn't lead to a mere differential function but rather to complicated ordinary differential equations which is hardly solvable without using a decent software:( After all, this is real world! – user36790 Aug 14 '15 at 3:21 • @user36790 thanks. Sometimes even though you can't solve something "exactly" doesn't mean you can't gain some insight from approximations. I agree that adding drag and spin would be interesting and complicated. – Floris Aug 14 '15 at 3:24 • +1; Yes, you are right too at this point; then it reduces to mere a kinematic problem but still I would prefer a better software use to an ideal-world physics equations. Nevertheless the answer was really good:) – user36790 Aug 14 '15 at 3:27 • @Floris - Another factor is that your$v$depends on the initial angle above horizontal. I've seen$v(\theta) = v_h \cos \frac \theta 2$in a few places as a simple model, where$v_h$is the initial velocity for a horizontally-thrown ball and$\theta$is the initial elevation, but I haven't seen any scientific references regarding this. This coupled with drag and lift (Magnus effect) reduce the optimal throw angle to well below 45 degrees, where "optimal" means the angle that yields the maximum distance. – David Hammen Aug 14 '15 at 12:14 • In other words, your$\alpha$is the release angle. If you change your$v$to$v_h \cos\frac \alpha 2$you'll find that, ignoring drag and lift, the optimal release angle for maximum distance is about 39.86 degrees: wolfram alpha calculation . – David Hammen Aug 14 '15 at 12:43 A 45 degree angle gives you the maximum distance to the point where the ball first hits the ground. That's not where a baseball will stop. If you hit the ball at a lower angle, more of the total speed will be in the horizontal direction, and the ball will still continue with a fairly high horizontal speed after bouncing. Thus, the point where friction finally stops the ball may be quite a bit further when you hit it at a low angle. If you throw a ball at 60mph at 0 degrees, the horizontal velocity is 60mph. However, if you throw it at a 45 degree angle, the horizontal velocity becomes$\sqrt{60^2/2}\$ mph, or about 42.4mph:

Thus, on a frictionless surface with no wind resistance, throwing it horizontally will get it to the base the fastest no matter how far it is away.

In the real world we have wind resistance and friction, so each bounce will slow the ball depending on the surface and ball-spin, and wind direction/speed will also come into play. We also have to worry about not having a bounce too close to your teammate to avoid errors. Here, a little loft will not slow the initial horizontal speed much and it will cause less bounces, so usually it is a good trade-off.

It's tough to cover all the bases though, so get out there and practice some different situations to figure out the best throwing angles in each. You could even get a stop-watch and time each other's throws in different situations to find the best angle that gets it there with the accuracy you desire. Have fun :)