so first, sorry this is again a little late. Mostly it's just too many things to do and not enough time to write.
I will have the math heavy version of equations of motion up tomorrow. It's just taken quite some time to write as I can't simply post equations I have to write them then turn them into pictures then host them.
This week, we will discuss what happens once you move out of linear motion.
Let's do a little experiment, let's say we have 2 friends (we'll call them Bob and Tom) and they are throwing snowballs each other. Now Bob and Tom have been going at it for quite some time and neither have managed to hit each other. Bob is down to his last snowball and desperately wants to land one on his friend. So how can we guarantee a hit for Bob?
First a refresher on the equations of motion we have:
Next a glossary of terms, as DK doesn't allow symbols and hosting symbols is well clunky at best and the most consistent critique from last week was more words; I have decided to set up a glossary of the terms I will use and what they mean:
theta -> always refers to angle. In this case the angle the projectile is thrown
V0 -> initial total velocity, note this is not necessarily equal to V0x or V0y. Always remember that you have to account for the fact that V0 is a vector and must be turned into it's components.
V0x -> this is the x component of V0, defined as cos(theta) times V0 (cos(theta) * V0) )
V0y -> this is the y component of V0, defined as sin(theta) times V0 (sin(theta)*V0 )
s0 -> initial position, in general this is defined as zero
s -> final position
t -> time elapsed
V -> velocity at any given point in time, technically a vector but I will discuss in more detail later
a -> acceleration, as we are considering the simplest of cases this generally refers the the gravity of the earth (-9.8 meters per second squared)
t(total) -> the total time elapsed as the object travels though the air
t(peak) -> time for an object to reach the maximum height during it's flight
sin-1 -> the inverse sine
Refer back to the glossary as I explain but initially it probably will not make much sense.
Let's talk about what we need to know to give Bob what he needs. Believe it or not we don't need much more then what I covered last week in Equations of Motion, all that we need to do is take that knowledge and add it to what I talked about in Vectors and Scalars.
Essentially velocity is a vector; in linear motion I ignored this because I did not want to complicate matters. More importantly though there is no need to discuss things in terms of vectors unless you really have to. Now I could do this in terms of pure math, but I am not if for no other reason then I promised not to and it is my hope that by using concrete numbers I can explain in a way that is more understanding. So let us start to get some facts, let us say that Bob throws his snowballs at a constant velocity of 50 meters per second (50 m/s). So now we have to break this vector into it's components which is actually quite simple. See all we need to do is take V0 and multiply it by cos(theta) to give us V0x. To get V0y we take V0 and multiply it by sin(theta). Why are we using these trig functions? Eh let's just say that there are mathematical reasons but that it is enough to now that cos(theta) times V0 will always give the x component and sin(theta) times V0 will always give the y component. That said, for those with a mathematical background the reasoning should be clear from just looking at the picture to the right.
So we've broken down V0 into it's components, but what does that give us? Well again we start to intrude into math, but basically because we can separate the components of the velocity we can treat dimension separately. This means that those equations above can be used to represent each dimension individually. Thus for example the first equation becomes: the y component of the velocity is equal to the y component of the initial velocity plus the acceleration times the time elapsed. ( Vy = V0y + a * t). This might not seem like much but trust me when I say that this result is what makes life very easy for calculating most ballistic trajectories.
This realization immediately yields some interesting and powerful results. For instance, given that I am keeping this simple we ignore wind resistance and thus have no acceleration in the x direction(and for ballistic motion generally this is a completely valid assumption) which means that x component of the velocity is just equal to the v component of the initial velocity ( Vx = V0x). Having no acceleration in the x component also means that the final position on the x axis (this is routinely called the range) is equal to x component of the initial velocity times the total time elapsed as the object travels though the air ( x = V0x * t(total).
This puts us very very close to what we are looking for, as we can find the x component of the velocity; it is equal to the initial velocity times cos(theta) we just need to figure out how to calculate the total time elapsed. In order to do this, I am going to invoke a trick I used last week.
Namely we know that the y component of the velocity is zero when the object reaches the height of it's flight ( V0y = 0) and we also know that the object takes just as long to go up as it does to go down.
Thus we will take the first equation and put in what we know. This gives us that the y component of the velocity (which is zero) equals the y component of the initial velocity plus the acceleration ( which is a constant - 9.8 meters per second squared) times the time required to reach the peak. ( 0 = V0y + ( - 9.8 m.s^2) * t(peak) ).
From there it's a simple exercise in algebra to solve for the time to reach the peak height. The time to reach peak height is the y component of the initial velocity divided by acceleration ( t(peak) = V0y / (9.8 m/s^2) ). From there we just multiply our result by 2 and we now have the total time the projectile is in the air ( t(total) = 2 * V0y / (9.8 m/s^2) ).
From here the end is insight all that remains is a lot of math. Given that the math heavy version of linear physics was delayed I am going to do the math results there and not here.
Thus the final answer we get is the angle we need is equal to one half times the inverse sine of the distance between Bob and Tom times 9.8 m/s^2 divided the square of the initial velocity. ( theta = .5 * sin-1( d * 9.8 / V0^2) )
So for those interested, if Bob and Tom are 150 meters apart, what is the angle Bob must throw his last snowball at if he throws it at 50 meters per second? I'll write the answer in the my tip jar.