Consider some abstract cases of pure mathematics. Begin with definitions. Velocity is the rate of change of position and acceleration is the rate of change of velocity. So velocity is meters per second and acceleration is velocity per second (or meters per second per second). Suppose constant acceleration. Using calculus you can derive some equations to describe one-dimensional motion:
- v = v_i + at
- x = x_i + v_i t + 1/2 at^2
The first equation gives you the object's velocity at any time. The second gives you its position. The _i means some initial condition. Here the object can only go back and forth on a single axis graph, and we try to set up so that x_i is equal to zero (the origin) to make things even simpler.
Note that according to Newton's laws of motion only forces create accelerations. This means the net external forces on the object result in only the acceleration in the kinematic equations of motion just listed. It will either speed up or slow down the object.
Suppose that the direction of an object launched from some fixed surface is said to be up and up is said to be positive; which means down is negative. Suppose that acceleration is -10 meters per second per second and that initial velocity is 3300 meters per second. The object is going to slow to a complete stop at some peak height during some interval of time, and then speed up in the reverse direction and return to the surface in the very same interval of time. Clearly where v = 0 meters per second is a special position. In fact, it's precisely half of the object's motion under consideration. Indeed, this is the point of symmetry for the motion. The rate of acceleration and the time interval for each half of the motion are precisely the same except the direction has flipped. If an object takes 2 seconds to reach the peak, it will take 2 seconds to fall back down.
This is how bullet drop is determined. Say that the source of acceleration is that of a gravity field. A bullet fired from a height of 11 feet or dropped from 11 feet will strike the ground at precisely the same time according to the equations. This has been experimentally verified. It's not all that difficult to test by oneself with some marbles and a table.
Now forces exert themselves in just one direction. Newton's third law tells us that all real forces come in pairs; each force is the same size but point in opposite directions. This is why there is no such things as centrifugal forces; it has no force pair. What is really happening when you go in circles is that Newton's first law is being proven. Anyways, the force of gravity, therefore, is in just one direction. On planets and moons that's down; it's attracting objects to the centre of the body. The force pair means the object is attracting the body towards its centre with the same size force, just in the opposite direction.
Sticking with our bullet, suppose that it's fired horizontally. This is gives the bullet's motion a second dimension. On a Cartesian graph we would like for the initial x and initial y values to be zero (the graph's origin) for simplicity; that's not always possible.
We can analyse its vertical motion; in fact, we have already said how long it will take to strike the ground. There are no forces exerting horizontally upon the bullet in this scenario because gravity is only vertical. Therefore, for this component of the bullet's motion, acceleration is zero. Newton's first law tells us that the bullet will continue in this way until some net external forces changes its motion (such as slamming into a target or intersecting with the ground). The point is that we use the same equations. They become significantly simplified when a = 0.
But what happens when the bullet is fired at some angle above the horizontal? We can still treat this two-dimensional motion as two one-dimensional analyses: one for the vertical and one for the horizontal. Clearly we must know the initial/muzzle velocity and the angle. In order to separate the two dimensions, we make use of right triangles. If we use the initial velocity as the hypotenuse, then we may multiply initial velocity by the cosine of the launch angle to know the horizontal velocity and we may multiply the initial velocity by the sine of the launch angle to know the vertical velocity. Using zero acceleration for the horizontal dimension we plug in the new initial velocity and solve the equations. We repeat the same for the vertical using the acceleration of gravity. If we need to recombine them, we may use the Theorem of Pythagoras.
In other words. For horizontal motion, use v_i cos(θ) instead of just v_i. For vertical motion, use v_i sin(θ) instead of just v_i. It is helpful to rewrite the equations for each dimension. Note that the angle θ is given in degrees not radians; be sure to set your calculator accordingly if you're using one.
In this way we can describe the most important elements of a trajectory for a given gravity field.
Whether through experimentation or familiarity with maths, some angles have noteworthy properties. At 45 degrees for a given muzzle velocity, the trajectory has the greatest range. At 30 and 60 degrees for a given muzzle velocity, the range of the trajectory is the same but shape is quite different (in fact, this is true of any complimentary angles). One way to see this is to get a calculator (or do it by hand, but calculators are faster -- see the monster at the end of this post for a link) and look at the boundary conditions: 0, 1, 29, 30, 31, 44, 45, 46, 59, 60, 61, 89, and 90 degrees.
How long would it take for our bullet dropped from 11 feet to fall to the surface of the Moon? We want to solve for time. We know its initial velocity (which is 0 m/s) and we know its acceleration (which is 1.667 m/s/s). We know it's initial x is 3.35 meters and its final x (the left side of the position equation) is 0 meters. We can plug all this into the equation, simplify, and solve for t. We may calculate the velocity of the bullet as it lands, too. If we fire the bullet at an angle, we can describe its trajectory in the same way as we did on Earth. In the case of the Moon where there is practically no atmosphere the results are going to be quite accurate. Indeed, an Apollo astronaut dropped a feather and a hammer while walking on the Moon and they fell at the same acceleration and landed at the same time.
One final comment for long range shots: the curvature of the planet or the moon has been completely ignored.
Are there other forces in reality? On Earth and other places where there is an atmosphere, the shape of the trajectory (which is a parabola in the pure case) can be deformed by other forces creating accelerations on the bullet. These forces are impossible to calculate outside of tightly controlled conditions and so their effect is impossible to calculate. But some general remarks should give some idea of what they can do to trajectories. Wind will exert force on the bullet and can come from many directions. Drag force for fast moving objects in atmosphere such as a bullet on Earth has several factors that go into its calculation; although the direction is always opposed to the motion of the bullet. The terms are the drag coefficient (which is close to .5 for spheres and for oddly shaped things it can get as high as 2); the air density which is determined by things like altitude, humidity, and air temperature; the cross-sectional surface area of the object the drag force exerts against; and the object's velocity squared. Friction is another force that will oppose the bullet's motion. It comes from contact with the bullet and atmosphere. In the case of larger objects, this can be a problem. Airplane designers slim down the vehicle to reduce drag but this elongates the vehicle which can increase friction (more surface area, more friction). Well, we can see the general overall effect in Mack's graphic where horizontal velocity compresses as we've added a negative acceleration into the system.
The following website has some useful trajectory calculators with some explanations of what is being calculated. It makes things easier. Trajectories can get quite tedious as there are often several steps to complete before arriving at the final answer. Calculators do the hard work. Although after some practice doing them by hand, they do become easier.
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
