The Angle of the Projectile

[asianboy]

It's been a long while since i have taken AP Physics in highschool, so I need some help.

I know that this is basic stuff, but I have bad memory

Say that an arrow has been shot at whatever angle. How do I calculate the angle of the arrow during the impact of the target at whatever range?

[Swagger]

you have to consider the speed at it was launched, the gravity and if you want something more specific, the wind resistance and other factors


i don't know the formula though

[Timoleon of Korinthos]

Assuming that the friction between air and arrow is minimal and thus can be neglected, then the horizontal velocity of the arrow remains the same throughout its course. If the velocity of the arrow at the moment it is shot is U, then the horizontal velocity is Ucosφ, where φ is the angle of shot (the reference being the horizontal level). The initial perpendicular velocity of the arrow is Usinφ. Let's say H is the height from which the arrow is being shot, namely the chest of the archer if he is shooting in a flat field or the latitude and the heigth of his chest if he is shooting from the edge of a cliff towards a flat valley.

Imagine a xy system of axes like this |__
From Netwon's (third?) law -mg=ma you get a=-g, a simple differential equation:

d(dy/dt)/dt=-g -> dy/dt=-g*t+C1 -> y(t)=-(1/2)*g*t^2+C1*t+C2 where y is the distance between the arrow and the ground, and C1,C2 are unknown constants.

This differential equation needs two initial boundary conditions in order to be solved(=define these constants).
As described above, the first one is y(0)=H and the second one dy/dt (0)= Usinφ

So you get y(t) = -0.5*g*t^2 + U*sinφ*t + H

The condition for the arrow hitting the ground is y(t)=0 => -0.5*g*t^2+U*sinφ*t+H=0

Solving this equation will provide you with two roots, one of them being negative and the other positive. You select the positive one, since time starts to count at the moment the archer takes the shot and negative time in this case would denote a moment before the starting point. Let's symbolise this root with tw

The perpendicular velocity at that moment will be dy/dt (tw) = -g*tw + U*sinφ

So now, you know both the horizontal and the perpendicular velocity and can thus find the angle which the arrow hits the ground throught their tangent. Taking the ground (=x axis) as the level of reference you get

θ = arctan[ (|-g*tw + U*sinφ|)/(U*cosφ)] (The | | is needed because of the way the xy axis and θ have been defined)

Picture it somehow like this:
arrow -> \
______θ\ ___

ground



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Edit: I have been thinking, you may want to find a general solution, meaning a solution for ordinary terrain, which is rarely flat.

In this case you need to know or find a function h(x) whose graph would match the form of the terrain as the distance between archer and arrow progresses, namely the height of the ground's surface in reference with a level you have chosen; you could choose the the sea level for example. (finding this h(x) is done through numerical analysis)
In order to relate this function with y(t) you have to find a relation between t and x. From the arrow's horizontal motion you get x=U*cosφ*t
Based on this you can transoform the function h(x) into h(t) by replacing every x in it with U*cosφ*t
If we define the height of the archers chest (point of reference being his feet) as h', then the equation you get is essentially the same as before, only the last part changes a bit, because now y(0) = h(0) + h':

y(t) = -0.5*g*t^2 + U*sinφ*t + h(0)+h'

Important: In this case the condition for the arrow to hit the ground is y(t)= h(t)
(I can't explain why this is this without a graph, but imagine that in this case the function y(t) is constructed as to show the vertical distance between the arrow and the level of reference you chose for the function h(t); so essentially it incorporates h(t), which shows the vertical distance of the ground's surface from the same level of reference. Thus, setting y(t)=h(t) means that every other factor that contributes in the function y(t) has been nullified, in other words that the arrow's orbit meets the ground's surface)

This equation, depending on the grade of h(t) is likely to give many roots. If so, the root tw you have to select is the smallest postive number, since once the arrow hits the ground the phenomenon ends. And from that point onwards you solve the problem exactly as before, except that your final angle θ will not be counted from the ground but from the horizontal level.

If you want to count it from the ground you will have to find h(tw) and then find the x which makes h(x)=h(tw). If there are a lot of roots, again you choose the smallest positive figure. Name it xw. Then you have to calculate arctan[dh/dx (xw)], which is the angle of the terrain's form at that point in reference with the horizontal level. Add this figure to the angle θ you have already calculated, and you get the angle counted from the ground's surface.
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Edit 2: Hey, I made a graph...kind of