A projectile is any object that, once set into motion, has no forces acting upon it other than gravity (and air resistance, though we ignore it whenever possible).
A projectile will follow a parabolic path until it hits something (at which point another force acts on it, and it is no longer a projectile). The projectile’s path also depends on its initial trajectory – the direction in which it is moving. If it starts out moving upwards at an angle, it will level off, and then begin to fall ever more steeply towards the ground. If it starts moving on the level, it will fall ever more steeply towards the ground. If it is already angled downwards at the start, the projectile continues downwards, just at ever steeper angles. You probably notice a pattern here, because all three are the same path, just different portions of the parabola.
*
[The Math]
If you want to perform calculations for projectile motion, you need to know something about the starting and/or ending conditions. These may include:
Initial height (h1)
Final height (h2)
Time of flight (t)
Initial velocity (v1)
Final velocity (v2)
Initial angle with the horizontal (L)
Distance traveled horizontally (d)
Acceleration of gravity (a) – on Earth this is -9.81 m/s^2
Because projectile motion occurs in two dimensions, it requires a pair of equations to solve a projectile problem. One equation represents vertical motion (up and down, along the y-axis) and the other deals with horizontal motion (which you can think of as forward motion, along the x-axis).
If you know the initial velocity (v1) and angle (), you can break the initial velocity into x- and y- components, to use separately in the two equations.
The x-component, v1x is found as:
v1x = v1 * cos(L)
The y-component, v1y is found as:
v1y = v1 * sin(L)
From this point on, the x- and y- motion of the projectile are independent of one another, except for one thing – time. The projectile moves only during its time of flight, which is the same for each. As a result, time is the variable that will tie our two equations together.
There are a number of basic equations for linear motion, all of which can be applied to projectile motion. There are two that are most commonly useful. For motion along the y-axis, which involves the acceleration of gravity, the following equation is most frequently used:
(h2-h1) = (a*t^2)/2 + v1y*t
Note that this is a quadratic equation, and will often have two solutions for time. Remember that you can’t have a negative time. A zero time means that you will have gone nowhere, and unless your projectile runs into something, it will only hit the ground on the way down, so you want to choose the larger of the two solutions for time. (Yes, this is common sense, but many students forget about common sense when faced with math.)
Horizontal motion occurs at a constant velocity (unless you take air resistance into account, then you get a negative acceleration and a little more math). Neglecting the air resistance, as most physics problems will, you can use the equation:
d = v1x * t
Note that you can rearrange this to solve for time:
t = d/v1x
If you are fond of algebra, you can then plug this into the y-equation and get one big equation for projectile motion.
(h2-h1) = (a*[d/v1x]^2)/2 + v1y*[d/v1x]
Most students find they make fewer mistakes with smaller equations, so unless you’re confident in your skills, you should use the pair instead.
*
[EXAMPLE]
A sample problem would be to calculate how far away a cannonball would land if it were fired from a cannon that was on a ledge 5.00 meters above the ground with an initial velocity of 20.0 m/s at an angle of 30.0 degrees (measured from horizontal).
We know specifically from the question that:
h1 = 5.00 m
v1 = 20.0 m/s
= 30.0 degrees
Implied in the question is also:
h2 = 0 m
a = -9.81 m/s^2 (acceleration due to gravity on Earth)
Begin by finding the x- and y- velocities:
v1x = 20 * cos 30 = 17.32 m/s
v1y = 20 * sin 30 = 10.00 m/s
Calculate the time spent in the air:
(h2-h1) = (a*t^2)/2 + v1y*t
(0-5) = (-9.81*t^2)/2 + 10*t
t = 2.454 seconds (the negative solution, -0.415 seconds, is discarded)
Calculate the distance traveled horizontally in that time:
d = v1x * t
d = 17.32 * 2.454
d = 42.5 meters
Obviously this was a weak cannon blast, so why don’t you go ahead and try it again, but using a cannon fired from ground height, at an angle of 60.0 degrees, with an initial velocity of 52.0 m/s. If you get 233 meters, you’ve got it!
*
[Other Equations]
Not all projectile problems will have the same given information, so you may sometimes find you need to use other equations to reach the final solution. Here’s your selection of linear motion equations. Between them, you’ll be able to solve it all. They can be applied in either direction, x or y, so rather than specifying height or distance, I will just use:
x1 = initial position
x2 = final position
a = acceleration
t = time
v1 = initial velocity
v2 = final velocity
v = constant velocity
–
x2 = v*t + x1
v2 = a*t + v1
x2 = (a*t^2)/2 + v*t + x1
v2^2 = v1^2 – 2a*(x2-x1)
x2 – x1 = t*(v1 + v2)/2
