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I am preparing a physics course for high school about projectile motions. If a projectile moves with initial velocity $v_0$ in the gravitational field of the earth, the equation

$$ s(t) = \frac{1}{2} g t^2 + v_0 t $$

holds, where $s(t)$ is the travelled distance at time $t$. Now I am looking for real world applications of this equation (or the corresponding equation for the velocity).

More specifically I don't want any problems where the equation is somehow artificially embedded into a real world situation, for example

"you fire the projectile of a signal pistol with initial velocity $v_0=...$, at which height is the projectile at time $t=...$, what is the maximal height..." projectile signal pistol In that example it is not clear why you know the initial velocity or why you want to calculate the maximal height (indeed, I think in most cases the manufacturer writes the maximal height into the manual, but then why should you be interested in calculating the initial velocity?)

So the point is, in the problems/examples I am looking for, it should be clear, why one has the input data and why one wants to calculate other things using the equation of motion.

Please stick to one problem/example per answer. Further references for the context of the example would be nice.

Edit: I should make clear, that I want examples where $v_0 \neq 0$. I am only intersted in the upward - downward -case.

One example I was thinking of, was that of a vulcano that ejects stones. However I don't know much about vulcanos, so I don't know which initial data are known, and what people want so calculate...

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    $\begingroup$ Hi Julia, and welcome to Physics Stack Exchange! On one hand, I really like the way you've written this question. But we've recently had a discussion on our meta site about questions like this which are looking for example problems, and the community seemed to support the idea that they are not really appropriate for this site... but perhaps the fact that you have explained in detail why the usual problems are insufficient will be enough to make this a good question here. $\endgroup$
    – David Z
    Commented Dec 11, 2011 at 3:01
  • $\begingroup$ There are hardly any real world problems where a projectile is thrown up- or downwards. I would recommend you extend your question a bit to allow throwing at an angle, there are lots of interesting examples available. The discussion on the meta site is concerned with the list format of the answers but I think this question has the potential to get insightful answers, even if there is not a single "right" one. $\endgroup$
    – Alexander
    Commented Dec 11, 2011 at 15:42
  • $\begingroup$ @Alexander Yes, it is clear, that it be a lot more easier to find examples where throwing at an angle is allowed. But that's the point of my question, I asked it, because it seems to be difficult to find good applications of the upward - downward case... $\endgroup$
    – Julia
    Commented Dec 11, 2011 at 17:59

8 Answers 8

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Another one:

A magician is locked in a wooden chest and is fired straight up from a very powerful cannon. He has a parachute to land safely, but it won't work if it's opened less than 25 meters from the ground. The magician tested the cannon (with a crash dummy, of course), and it can launch him up to 50 meters high. How much time will he have to escape from the chest?

A different version, less contrived but a little more difficult:

...The magician tested the cannon (with a crash dummy, of course), and it takes the dummy 6 seconds to crash bask into the ground. How much time will he have to escape from the chest?

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If the initial velocity $v_0$ is allowed to be zero, here is one:

You are standing next to a very deep hole, and wonder what is the depth $h$? You drop a stone into the hole, and hear a delayed sound (of the stone hitting the bottom) after time $T$ on your stopwatch. Given the gravitational constant $g$, given the speed of sound $c$, and by ignoring air-resistance on the stone, calculate the depth $h$.

Solution:

$$ h ~=~ cT + \frac{c^2}{g}\left[ 1- \sqrt{1+ \frac{2gT}{c}}\right].$$

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  • $\begingroup$ Yes, that's a pretty nice example of the free fall, however I want only cases where $v_0 \neq 0$. I will edit my question to make this point clear. $\endgroup$
    – Julia
    Commented Dec 11, 2011 at 17:57
  • $\begingroup$ Perhaps use a symbol other than c for the speed of sound ;) $\endgroup$
    – tyblu
    Commented May 1, 2012 at 7:04
  • $\begingroup$ An anonymous reader suggested to take into account the difference in altitude between your ears and your hand. Let us for simplicity assume that these two body parts are held at the same altitude, and $h$ is measured relative to that altitude. $\endgroup$
    – Qmechanic ♦
    Commented Feb 7, 2013 at 16:39
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Not related to guns or projectiles I have recently used the 2D version of this equation to make sure oil is squirted into an engine at the correct location using an "oil jet". If the oil did not land in the correct location, then the engine components will overheat and eventually fail (bad). So in a design sense, the equation allowed me to find the correct position for the oil squirters. The oil comes out of the orifice in a steady stream, with known exit velocity based on the oil pressure behind the hole.

The reason velocity has to be known is because of Newtons Laws of motion which help us take an frozen instant in time and calculate accelerations from positions and velocities. In its core, determinism means if I know the state now, I can predict it in the near future. This is all it is done with the projectile equations.

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  • $\begingroup$ Though it didn't answer the original question, the context of your example is interesting and I would like to create a problem out of it. Since I don't know the your context in detail, it would be nice if you could add some more details, for example a sketch of the engine and the path the oil should travel as well as some realistic numbers for this problem. $\endgroup$
    – Julia
    Commented Dec 2, 2017 at 20:46
  • $\begingroup$ Google "piston oil squirters" $\endgroup$
    – John Alexiou
    Commented Dec 2, 2017 at 22:00
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This one could work (and it's based on real events):

You are at a friend's apartment and you are about to leave. He's very lazy, so he gives you the keys of the door at the ground floor. You get out of the building and you try to toss the keys back to him through an open window (or at his balcony), that's about 9 meters high. How fast should you throw the keys?

Or, if a small bending of the rules is allowed:

...that's about 9 meters high. You've played throwing rocks before, so you know that you can't toss objects similar to the keys further than about 30 meters. Can you throw that high? What's the highest floor that you could throw the keys at?

Of course, check that the numbers work for you.

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A jumping astronaut is an example that might be interesting.

I always found it very exciting to watch documentaries about the Apollo missions and watch how the astronauts are jumping around on the moon. From our perspective it looks almost cheerful, as they can jump higher and the whole movement looks therefore much slower.
Considering that the inertia is the same whether your are on the moon or not it seems not far fetched that the initial velocity from the astronauts does not really depend much on gravity and therefore the maximum height and time that the astronaut needs to complete the jump is mainly determined by the moon's lower gravity.

So for an example the students could perform small jumps in the classroom and measure the time until they hit the ground again (with a backpack to create similar conditions). Now compare this time with some videos from the Apollo missions and try to determine the change in $g$ by comparing the different times.

While this will not be precise I would guess you should end up in the right ballpark of $$g_{moon} = 0.167 \cdot g_{earth} $$

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Here's one example that almost fits your qualifications: one time on Mythbusters they investigated the behavior of bullets fired straight up in the air, or what people would consider "straight up" which in practice means at a small angle away from truly vertical. The main thrust of the investigation was to determine how far away a bullet would travel horizontally, in other words to determine the size of the "danger zone" in which bystanders are liable to be hit by the falling bullet. And of course, to do that you need to calculate the flight time using the muzzle velocity of the bullet. It's probably safe to assume some small angle like $3^\circ$.

I forgot to mention at first: this wouldn't be a realistic calculation because air resistance is very significant in reality. But you can still use it as a kinematic example.

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    $\begingroup$ not a very good example because air drag is not neglectable. In practice (i.e. with drag) a high powered rifle bullet will reach about 10,000 feet. But if you neglect drag, you would calculate a 2700ft/sec (823m/s) bullet to reach over 100,000 feet. Actually let me have another go at it - this is actually a VERY good example of learning where it is OK to neglect air drag and where it is not. $\endgroup$
    – Daniel Chisholm
    Commented Dec 12, 2011 at 13:50
  • $\begingroup$ Oh yeah, I was going to mention that. $\endgroup$
    – David Z
    Commented Dec 12, 2011 at 20:24
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Since the previous example was biased, hence i am adding another example,

Suppose there is a hot air baloon at a point a,b in 2d world. and a gun which is kept at origin is designed to hit the baloon.

then find the minimum velocity of the bullet of gun,(which follows a projectile path) so that it hits the hot air baloon.

{This is problem of inclined plane projectile,analogy}

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(a bit like the oil jet answer..) What about a fountain which directs a jet of water essentially straight up?

Pros: an everyday real-world non-contrived example

Cons: - it involves a stream rather than a particle. The math of following a conceptual "fluid particle" is the same but it would probably be more satisfying to have an example that involves following a point mass

  • you wouldn't directly know the speed of the jet at the base of the fountain but it would be obviously calculable (fluid mechanics, pressure drop is 0.5 * rho * v^2).
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