I wrote this post a couple of years back on my old website, which has long gone. I thought it could do with a shameless repost.
Hawkeye should be as far away from danger as possible!
Hawkeye gets a bad press. One of the 'normal' humans in the Avengers. No superpowers, except he's really good at darts. Which unless he's in a showdown with Phil 'The Power' Taylor, isn't probably very helpful to anyone. If I was him, I'd want to be shooting that bow from as far away as possible from Thanos, Ultron, or whoever. So what's the best angle he should shoot his bow at, in order to avoid danger?
Once Hawkeye shoots the arrow, it becomes a projectile. And to work out how Hawkeye can get maximum distance, we're going to start by looking at how fast the arrow is going, or its velocity.
A projectile is basically any object that has been dropped or thrown (projected). It will continue to move by its own inertia and is influenced only by the pull of gravity. It will move an a parabola, a symmetrical curve (as long as we ignore air resistance and drag, which we will).
Because Hawkeye will most likely have to launch his bow at an angle to the horizontal, it will have a speed (or more correctly velocity) that can be broken into two components. A vertical velocity, and a horizontal velocity.
Both the horizontal component and vertical component can be considered independent from each other.
Now I'm about to use a term called 'cosθ', and in a bit I'm going to use a term called 'sinθ', which unless you are a maths nerd like me will result in painful flashbacks to maths lessons, and you reaching for a paper bag to breathe heavily into. Just hold on, it will be explained hopefully as painlessly as possible.
The magnitude of the velocities are represented to scale by the length of their arrows. The longer the arrow, the greater the velocity. When combined with the initial velocity, all three lengths can be combined to make a right angled triangle (a triangle that has one angle of 90°).
Because the three velocities together form a triangle when drawn to scale it means that we can use trigonometry to find the magnitude of the horizontal and vertical velocities. And that's it, you can stop breathing heavily into that paper bag now.
Horizontally, once the arrow has left the bow, there are no external forces acting on it. Meaning that from the left of from the right there is nothing pushing or pulling on it. Because of this the velocity horizontally will be the same magnitude (size) throughout the journey of the arrow, as there is nothing pushing on the arrow, making it slow down or speed up.
Because gravity is always pulling directly downwards on the arrow, it will cause the arrow to change its velocity vertically. Once launched, the arrow's vertical velocity will gradually slow down until it reaches its highest point and will have a velocity of zero. Once it reaches it's highest point it will gain velocity in the opposite direction and start to speed up. Gravity will cause the arrow to decelerate when it is going upwards, and acclerate when it is going downwards at a rate of 9.81 metres per second, every second. In any equation using the acceleration due to gravity, we label that value 'g'
This means that if it was going up at a vertical velocity of say 30 metres per second (m/s), after a second the vertical velocity would be 20.19 m/s (30 - 9.81). After two seconds the vertical velocity would be 10.38 m/s (30 - 9.81 - 9.81).
It also means that after the arrow reaches its highest point and as it starts to 'fall' the arrow will gain a vertical velocity of 9.8 m/s every second that it is falling.
Now instead of using - 9.81 - 9.81 - 9.81 - ... for every second of time, it's easier to multiply 'g' by the number of seconds and subtract that from the starting vertical velocity.
We're going to use these two equations as a starting point to work out everything else. How long the arrow is in the air, distance traveled and finally launch angle.
How far does the arrow go?
We're going to use the two equations above to work out a few things, first off we are going to figure out what the distance the arrow would travel horizontally in terms of the arrow's velocity. The next time you're in a car look at the speedometer, you'll see the units in mph (miles per hour) or kph or km/h (kilometres per hour). This means that:
So we have range in terms of launch angle and time taken. We have two unknowns, so how can we 'get rid of time'?
What do we know about the time of flight?
We know that if there is no air resistance, the path of the arrow is symmetrical. That means that when it is at its highest point, the arrow is halfway through its journey time. We also know that at its highest point, the vertical velocity is 0. So let's go back to the equation for vertical velocity and see if we can relate it to time in any way:
Now this 't' in the equation above is the time taken for the arrow to reach its highest point. This is half the amount of time the arrow spends in the air overall.
We can put this expression for time in the air into our equation for the range of the arrow.
If Hawkeye always fires his arrow at the same launch speed no matter what angle he aims at we can label that a constant (it doesn't change). Gravity 'g' also doesn't change. This means that the only thing that can change is sinθ and cosθ.
Every other value in the equation (velocity, gravity...) has a constant value, and they will be multiplied by sinθ multiplied by cosθ. This means that to get the biggest range, we need sinθ multiplied by cosθ to produce the biggest value possible.
Crunching the numbers
We find the biggest value of sinθ multiplied by cosθ by trial and error. But I love a good spreadsheet, so I decided to plot the values of sinθ multiplied by cosθ for all angles from 0 degrees (completely horizontal) to 90 degrees (shooting straight up), to see which would give us the biggest value.
When we look at the chart, we see that the biggest value of sinθ multiplied by cosθ is when θ is 45 degrees.
This that for Hawkeye to be as far away from villains with actual real powers he needs to shoot his arrow with an angle of 45 degrees. Or he could just let Hulk smash them, which is what I'd do in his position.