# Thread: How far will this object travel?

1. It's been two decades since I last did any physics calculations, so I'm hoping somebody can help me with the following thing I'm trying to work out. I tried using some online physics calculators, but can't get it right.

In the athletics (track and field) event shot put, some rule changes have made the shot 9 grams heavier than it used to be, but without any changes to records etc. The question therefore arises - what difference does 9 grams really make?

The following parameters can be used. The "new" shot, that weighs 7265 grams, is used. It is released from a height of 2,20 meters above a perfectly flat competition area at an angle of 40 degrees from horizontal and hits the ground 20.00 meters from the point of release. If everything stays exactly the same, but the "old" shot, that weighs 7256 grams, is used, how far will it travel before hitting the ground? Wind resistance can be ignored.

For in case it's needed, here are some of my amateur ideas about how this calculation should be done.
1) The speed of the shot in the direction of movement at release should be calculated.
2) The amount of force needed to achieve that speed needs to be calculated (would the distance over which the force is applied make a difference? If it does, I can supply some typical values)
3) The speed at release if the same force was applied to the lighter shot must be calculated.
4) The distance of travel of the lighter shot must be calculated.

Thanks!  2.

3. 20cm  4. Here's a hint : write down the general expression for the trajectory of the shot put. You know the initial angle and height, and the distance until it hits the ground as well as the mass. From that you can calculate the initial speed of the shot put.
Now take the general expression and calculate the distance with the heavier shot put - all parameters are the same, except the mass and the resulting distance. I don't think you are required to calculate any forces, the way I understand the exercise is that the initial velocity is the same in both cases - if that wasn't the case there'd be little point in comparing the two trajectories.  5. Originally Posted by Geo 20cm
Nice guess. Certainly wrong though!  6. Originally Posted by Markus Hanke Here's a hint : write down the general expression for the trajectory of the shot put. You know the initial angle and height, and the distance until it hits the ground as well as the mass. From that you can calculate the initial speed of the shot put.
Now take the general expression and calculate the distance with the heavier shot put - all parameters are the same, except the mass and the resulting distance. I don't think you are required to calculate any forces, the way I understand the exercise is that the initial velocity is the same in both cases - if that wasn't the case there'd be little point in comparing the two trajectories.
Thanks for the suggestions Markus, but you're already losing me when talking about "general expression".

Also, I think you're probably wrong when saying that the initial velocity should be the same. First, the athlete puts the shot with absolutely the maximum force his body is capable of, so when using the heavier shot he will not be able to get it moving at exactly the same velocity. Second, if the different weight shots were released at the same velocity, wouldn't they travel exactly the same distance - The trajectory would be the same, irrespective of what they weigh?  7. Originally Posted by Gerdagewig  Originally Posted by Markus Hanke Here's a hint : write down the general expression for the trajectory of the shot put. You know the initial angle and height, and the distance until it hits the ground as well as the mass. From that you can calculate the initial speed of the shot put.
Now take the general expression and calculate the distance with the heavier shot put - all parameters are the same, except the mass and the resulting distance. I don't think you are required to calculate any forces, the way I understand the exercise is that the initial velocity is the same in both cases - if that wasn't the case there'd be little point in comparing the two trajectories.
Thanks for the suggestions Markus, but you're already losing me when talking about "general expression".

Also, I think you're probably wrong when saying that the initial velocity should be the same. First, the athlete puts the shot with absolutely the maximum force his body is capable of, so when using the heavier shot he will not be able to get it moving at exactly the same velocity. Second, if the different weight shots were released at the same velocity, wouldn't they travel exactly the same distance - The trajectory would be the same, irrespective of what they weigh?
That's probably correct. Separating the movement into horizontal and vertical components, both will accelerate at the same rate vertically and if you ignore wind resistance, both will have the same horizontal component as well. The variables are the increase of inertia of the heavier ball, meaning wind resistance will slow it down less, and the initial velocity the athlete will be able to release the weight at. We don't have figures on either of those variables.

In the end, the momentum the athlete can supply would remain the same, so we should be able to work out what velocity the athlete will be able to release the lighter ball at and subsequently how far it should be able to travel.

Why did they change the mass of the ball in any case?  8. Originally Posted by KALSTER That's probably correct. Separating the movement into horizontal and vertical components, both will accelerate at the same rate vertically and if you ignore wind resistance, both will have the same horizontal component as well. The variables are the increase of inertia of the heavier ball, meaning wind resistance will slow it down less, and the initial velocity the athlete will be able to release the weight at. We don't have figures on either of those variables.

Why did they change the mass of the ball in any case?
If we leave the inertia and wind resistance out of the equation (I think the difference would be negligible), it only leaves the difference in initial velocity. That's basically exactly what I'm asking for above. I'm hoping somebody can use the parameters I supplied to work out what the difference in velocity and the resultant difference in distance traveled would be.

The ball used to be standardized at 16 lb's (a standard artillery weight - the first shot put competitions was to see who could throw a cannonball the furthest), which works out to 7256 grams (or maybe 7257?). They then metricized it to 7260 grams, where it still is today. However, to allow for a little margin of error, a shot supplied by the organisers at the Olympics for instance must weigh at least 5 grams more = 7265 grams.  9. Using this trajectory calculator I get an initial velocity of 13.272 meters per second with the heavier shot.
Ballistic Trajectory (2-D) Calculator - Computes the maximum height, range, time to impact, and impact velocity of a ballistic projectile

I would expect that the kinetic energy should be equal in both cases. Kinetic energy is proportional to the mass and the square of the speed.
m1v1^2=m2v2^2
(v2/v1)^2=m1/m2=7265/7256=1.00124
v2/v1=sqrt(1.00124)=1.00062
v2=13.280
Plugging that into the calculator gives a range of 20.02 meters. So the lighter shot would get you an extra 20 millimeters.  10. Originally Posted by Harold14370 Using this trajectory calculator I get an initial velocity of 13.272 meters per second with the heavier shot.
Ballistic Trajectory (2-D) Calculator - Computes the maximum height, range, time to impact, and impact velocity of a ballistic projectile

I would expect that the kinetic energy should be equal in both cases. Kinetic energy is proportional to the mass and the square of the speed.
m1v1^2=m2v2^2
(v2/v1)^2=m1/m2=7265/7256=1.00124
v2/v1=sqrt(1.00124)=1.00062
v2=13.280
Plugging that into the calculator gives a range of 20.02 meters. So the lighter shot would get you an extra 20 millimeters.
Great! That's exactly what I was looking for! Thanks a lot Harold.

This could actually have a serious implication in competition. I know of cases where some of the shots supplied by the organizers were 30+ grams over the minimum weight (without the athletes knowing). Your calculations show that it could therefore travel at least 6 cm shorter than a shot with the correct weight. Sometimes 6 cm is the difference between gold and not even getting a medal!  11. Originally Posted by Gerdagewig  Originally Posted by Harold14370 Using this trajectory calculator I get an initial velocity of 13.272 meters per second with the heavier shot.
Ballistic Trajectory (2-D) Calculator - Computes the maximum height, range, time to impact, and impact velocity of a ballistic projectile

I would expect that the kinetic energy should be equal in both cases. Kinetic energy is proportional to the mass and the square of the speed.
m1v1^2=m2v2^2
(v2/v1)^2=m1/m2=7265/7256=1.00124
v2/v1=sqrt(1.00124)=1.00062
v2=13.280
Plugging that into the calculator gives a range of 20.02 meters. So the lighter shot would get you an extra 20 millimeters.
Great! That's exactly what I was looking for! Thanks a lot Harold.

This could actually have a serious implication in competition. I know of cases where some of the shots supplied by the organizers were 30+ grams over the minimum weight (without the athletes knowing). Your calculations show that it could therefore travel at least 6 cm shorter than a shot with the correct weight. Sometimes 6 cm is the difference between gold and not even getting a medal!
I presume everyone throws with those same balls then, or do you mean some had balls 30 grams heavier and others not?

Nice Harold, I managed to miss that one while Googling for a ballistics calculator.  12. Originally Posted by Harold14370 Using this trajectory calculator I get an initial velocity of 13.272 meters per second with the heavier shot.
Ballistic Trajectory (2-D) Calculator - Computes the maximum height, range, time to impact, and impact velocity of a ballistic projectile

I would expect that the kinetic energy should be equal in both cases. Kinetic energy is proportional to the mass and the square of the speed.
m1v1^2=m2v2^2
(v2/v1)^2=m1/m2=7265/7256=1.00124
v2/v1=sqrt(1.00124)=1.00062
v2=13.280
Plugging that into the calculator gives a range of 20.02 meters. So the lighter shot would get you an extra 20 millimeters.
I've just noticed an interesting thing on the calculator. For a shot putter who throws as described above, the optimal angle to release the ball is about 42 degrees, not 45 as commonly assumed. It is obviously so because of the fact that the ball is landing lower than it is released from, but this is generally overlooked. Studies have shown that top shot putters throw the shot at about 38 to 42 degrees, and it was always assumed that that was below the optimal angle of 45 degrees. It was speculated that the reason why athletes did it that way because it was the best compromise between getting the perfect angle and maintaining the horizontal speed that was built up during the initial movement through the circle before the final throw is made. The theory is that the initial horizontal velocity is steered upwards during the final phase of the throw, but that if the angle of change during this "steering" is too high, too much of the momentum in the projectile is lost. All of these calculations are of course not made consciously by those stupid fat guys but is just "felt" and learned over years of practice. This calculator now makes it clear that some guys are actually already throwing at the optimal angle and can't throw even further if they achieved 45 degrees (as is often assumed by coaches and athletes alike).  13. Originally Posted by KALSTER ]I presume everyone throws with those same balls then, or do you mean some had balls 30 grams heavier and others not?
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There is a range of shots supplied (at the Olympics it could be as many as 20 different shots) and the athletes just assume that they all weigh exactly the same. I know however that some of the shots at major competitions (probably not at the Olympics though) are much heavier than others. The reasons for such a wide range of shots is that they vary slightly in diameter (minimum 110mm and max 130mm), substance (iron, steel, copper) and finish (smooth, slight texture, painted or not etc).

Keeping the difference in diameter in mind, I at first thought it makes sense that it would be best to throw with the smallest possible shot, because you would gain 1 cm compared to doing the same throw with the biggest possible diameter. The reason is that the measurement is taken from the closest point of contact on the ground and if the centre of mass moves the same distance, the closest point of contact will be 1 cm further. However, this is wrong because the centre of mass would also be 1 cm closer to your hand at point of release. The only difference would be in wind resistance, which I think is negligible. Almost all throwers prefer the biggest possible shot, because it provides a flatter surface to exert the force on, which is more effective than on a more rounded surface.  14. Originally Posted by Gerdagewig
For a shot putter who throws as described above, the optimal angle to release the ball is about 42 degrees, not 45 as commonly assumed. It is obviously so because of the fact that the ball is landing lower than it is released from, but this is generally overlooked.
Not sure if this is accurate. The ball maintains it's horizontal speed all the way down. Not sure how you arrived at 42 degrees?

While Googling around, I also read that due to physiological variables, athletes can attain higher release velocities at different angles. While a certain angle might be ideal, it could be compensated for or overcome by being able to attain a higher initial velocity at other release angles. So an athlete that shoots at 38 degrees might actually lose distance if they tried it at the "optimal" 45 degrees. It might be possible to improve the angle by training muscles to function more optimally when the athlete shoots at closer to 45 degrees.

Quite interesting.   15. A thrower does however gain a centimeter or three when the shots are landing on a very hard surface (highly compacted soil or tartan for instance) as compared to a soft surface (wet-ish grass for instance). The reason for this is, as described above, that the measurement is taken form the nearest point of contact. On a hard surface the shot will make an imprint that is a few centimeters smaller in diameter than on a soft surface. If the centre of mass traveled the same distance, the throw on the hard surface would therefore be measured a bit further.  16. Originally Posted by KALSTER  Originally Posted by Gerdagewig
For a shot putter who throws as described above, the optimal angle to release the ball is about 42 degrees, not 45 as commonly assumed. It is obviously so because of the fact that the ball is landing lower than it is released from, but this is generally overlooked.
Not sure if this is accurate. The ball maintains it's horizontal speed all the way down. Not sure how you arrived at 42 degrees?

While Googling around, I also read that due to physiological variables, athletes can attain higher release velocities at different angles. While a certain angle might be ideal, it could be compensated for or overcome by being able to attain a higher initial velocity at other release angles. So an athlete that shoots at 38 degrees might actually lose distance if they tried it at the "optimal" 45 degrees. It might be possible to improve the angle by training muscles to function more optimally when the athlete shoots at closer to 45 degrees.

Quite interesting. I noticed the fact that 42 degrees is the best on the ballistic calculator (I always thought 45 degrees was optimal). After thinking about it and taking the difference in release and landing heights into consideration, it started making sense to me. Even though the horizontal speed stays the same, it would be in the air slightly longer at the lower angle because of coming in to land slightly shallower. The lower angle can thus produce a slightly longer throw because the shot is in the air slightly longer.

The physiological factors that might make it better to throw at a lower angle is what I referred to above when saying that a thrower must "steer" the implement upwards during the final phases of the throw. It certainly helps if the right muscles are strong, but the technique and bio-mechanics probably plays the biggest role and getting the best angle.  17. I was wrong. I just checked the flight time on the calculator and see that the shot would in fact be in the air longer when thrown at 45 degrees than at 42 degrees. I'm confused now.  18. After some more thought (which is difficult for me!) I think the following is the best way to look at this. At the lower angle the horizontal speed is higher than at 45 degrees. Even though it is in the air for a slightly shorter time, it therefore still covers a longer horizontal distance, especially in the last 2.2 vertical meters before it lands. The throw at 45 degrees had a longer absolute travel distance (along the path it took through the air), but not when measured just horizontally. This effect quickly disappears when lowering the angle further (to 38 degrees for instance) because the increased horizontal speed doesn't make up for the decreased flight time anymore. If the release and landing heights are the same, the best angle is 45 degrees, not 42 degrees.  19. Originally Posted by Gerdagewig I was wrong. I just checked the flight time on the calculator and see that the shot would in fact be in the air longer when thrown at 45 degrees than at 42 degrees. I'm confused now.
Me too. The angle the ball would go through below the release hight as it comes down would be the same as the release angle, if I am not mistaken. Doing the calculation at a downward angle of 45 degrees instead of upward should still be better than 42 degrees, as far as I can wrap my head around it. I have a feeling that had the ball been allowed to drop the same distance below the release hight it was able to travel above it, that the 45 degree angle might still win out. That means that there are different optimal angles for athletes of different hight as well. Confusing.

Edit:

If the release and landing heights are the same, the best angle is 45 degrees, not 42 degrees.
I see you arrived at the same conclusion.   20. Let's look at this carefully - if air resistance can be neglected, the trajectory of the shot is where I choose the origin of my coordinate system to be at the point where the shot leaves the athlete's hand. The first thing that is immediately obvious is that this trajectory does not directly depend on the mass of the shot, it only depends on the angle and speed at which it leaves the athlete's hand. Unfortunately we are not given any data as to how exactly the athlete accelerates the shot, so we can only use the very general second law of Newton and thus where F is the force the athlete applies to accelerate the shot, and m is its mass, and t the time the acceleration is applied for. We know none of these apart from the mass, so all we can do is assume that everything stays the same except the mass m - we see immediately that, if we increase the mass by a factor of 9/7256 as described in the OP, we will via the above relation decrease the resultant initial velocity by the same factor ( aotbe ), which makes sense since the athlete has to work slightly harder to accelerate the heavier shot, so if he uses the same force he gets a lower velocity.
Given all this we can plug in the numbers, and arrive at a result, since we have two equations for the two unknowns t and v in the trajectory vector above.
I will leave that bit up to yourself to do.

That means that there are different optimal angles for athletes of different hight as well. Confusing.
See the trajectory vector above. It is quite straightforward, given the right data.  21. Originally Posted by Markus Hanke Unfortunately we are not given any data as to how exactly the athlete accelerates the shot, so we can only use the very general second law of Newton I made the assumption that the kinetic energy imparted would be equal. I think this is reasonable, because the kinetic energy will be an integration of work done on the shot, which is an integral of force multiplied by incremental distance. The athlete will not appreciably change his throwing motion with a minor change in weight, and should be applying a very similar force at each point in the throwing motion.
In an extreme case, if the projectile were much lighter, it would be hard to apply as much force. Think of throwing a ping-pong ball. But I don't think a minor change in weight would alter it much.  22. Hi Markus and Harold. Thanks for your input on this question. I would certainly not have been able to do these calculations without your help. I think it's pretty clear that a difference of about 10 grams does in fact have a small measurable influence on the distance of the put. The discussion also made it clear that 45 degrees is not the optimal angle for putting a shot as almost every thrower assumes, but rather something like 42 degrees.  23. Originally Posted by Harold14370 I made the assumption that the kinetic energy imparted would be equal. I think this is reasonable, because the kinetic energy will be an integration of work done on the shot, which is an integral of force multiplied by incremental distance. The athlete will not appreciably change his throwing motion with a minor change in weight, and should be applying a very similar force at each point in the throwing motion.
In an extreme case, if the projectile were much lighter, it would be hard to apply as much force. Think of throwing a ping-pong ball. But I don't think a minor change in weight would alter it much.  24. Originally Posted by Gerdagewig Hi Markus and Harold. Thanks for your input on this question. I would certainly not have been able to do these calculations without your help. I think it's pretty clear that a difference of about 10 grams does in fact have a small measurable influence on the distance of the put. The discussion also made it clear that 45 degrees is not the optimal angle for putting a shot as almost every thrower assumes, but rather something like 42 degrees.
I agree and think you have a good grasp on the reason behind it. Does the 42 vs 45 degrees make a difference to the athlete, i.e., does he have a way of measuring it?  25. Originally Posted by Harold14370 . Does the 42 vs 45 degrees make a difference to the athlete, i.e., does he have a way of measuring it?
In the average training session there isn't really an accurate way of measuring it, but after watching literally millions of throws over decades of involvement in the sport, a top coach develops a very good eye for this and can immediately tell if a throw is slightly lower than normal. In fact, a top coach can even predict a low throw half a second or so before the implement is released, because certain technical mistakes early in the movement invariably leads to a low throw.

Those throwers who are lucky enough to have access to state of the art training facilities regularly go for video analysis. All aspects of the bio-mechanics of the throw are then analysed in detail. A primitive method of measuring the angle of the throw is to have the athlete throw into a net that hangs about 1.5 meters in front of the athlete. By carefully observing where the shot hits the net and where it was released from etc, a relatively accurate angle can be worked out.  26. Yes 2cm not 20cm. One order out.

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