I will show you how to solve pretty much any projectile motion problem you face. First,
we have to know what equations to use, then I will get into using a coordinate system. There
are 4 equations you will need. 1 will be for horizontal motion, while the other 3 will be
for vertical motion. The first equation, which is for horizontal motion, is this:
Here, x is the final displacement, or final location of the object in
the horizontal direction. X0 is the initial location and V0 is the horizontal component
of the velocity. One thing to remember is that velocity in the horizontal direction
is always constant because we don’t take into account wind resistance. T is time.
Now, the rest of these equations are for vertical motion. They are the following.
In these equations, vy is the final vertical component of the velocity of the object. Voy
is the initial vertical component of velocity. A is acceleration of gravity,
which is 9.81 m/s^2, or 32.2 ft/s^2. T is time. Y is the final location in the vertical
direction and y0 is the initial location. In most cases, you would only use the
first equation and the 3rd equation to solve problems. Let’s look at
some examples and see how we can solve them. In this example, it’s asking us to determine
the angle and speed the ball was thrown. Before anything, we need establish a coordinate system.
I almost always put the coordinate system at the location at which the object is launched at. I
establish that up is positive and to the right is positive. Where you place the coordinate system is
up to you, you can place it at the final location, or the initial location, but when you do that,
you have to remember to use proper positive and negative signs when finding out the distances
and velocities from where the coordinate system starts. That is our initial location. Let me
explain. In this example, I placed the coordinate system at the initial point. So at that location,
the horizontal and vertical distances are 0. Now, we need to write down what we know. This
part is important so that you won’t get confused with the positive and negative values. So let’s
first think about only the horizontal direction. Point A is at 0 m, since that’s where we placed
our coordinate system. Point B is to the right, so it’s 50 m away from the origin. The velocity
must be broken into 2 components, horizontal and vertical. The horizontal component is pointing
to the right, so again, it’s a positive value, which in this case is Vcos theta.
Now, we can write what we know for vertical components. Point A is at 0 m. Point B,
is at -1.2 m. This is where I said we need to be careful about our positive and negative values.
Notice how point B is below point A, in other words, it’s below the x-axis of our coordinate
system, which means it is a negative value. The vertical component of velocity as it’s
being thrown upwards is a positive value, which is Vsin theta. We also have gravity,
which is downwards in pretty much every case, so that’s -9.81 m/s^2. The question also says
it took the ball 2.5 seconds to strike the ground, so that’s our time value.
Once you write down everything you know, it’s time to write 2 equations.
We need 2 because the questions will almost always ask you to find 2 unknowns or a variation of 2
unknowns, in very rare situations, you would need 1 horizontal equation and 2 for vertical motion.
In this question, one will be for horizontal motion, and another one will be for vertical
motion. So let’s start off with horizontal motion. For that, we can use this equation.
Let’s substitute what we know into the equation. Final location is 50m, initial location is at 0m,
the horizontal component of velocity is Vcos theta, and time is 2.5 seconds. Now,
we can set that aside, and write another equation, this time, for vertical
motion. For that, we can use this equation. Let’s do the same as before, and substitute
what we know into this equation. The final location in the vertical direction is -1.2 m,
initial location is 0, the vertical component of velocity is v sin theta, gravity is -9.81m/s^2,
and time is 2.5 seconds. Now, we have 2 equations with 2 unknowns. You can solve them any way you
like. Once you solve the 2 equations, you will have your 2 answers. This question is fully solved
step by step, please check the description. Now, let’s look at another example.
In this question, a bag is dropped from a hot air balloon, and we have to figure out how long
it took to fall, and at what speed it strikes the ground. So first, we will establish a coordinate
system. I will place the coordinate system at the location of the bag. Let’s write down what
we know. The question says to assume the bag was released with the same velocity as the balloon,
so let’s keep that in mind too. For the horizontal direction, we have the starting location at 0 m.
The velocity in the horizontal direction is -20 km/h which is -5.556 m/s. It’s negative
because it’s to the left, and our coordinate system is set for right side to be positive. Now,
let’s look at the vertical direction. The bag starts off at a location of 0m, and then drops
-50 m, which is our final location. We also have velocity going upwards, so we have 12 km/h which
is 3.333 m/s. Don’t forget gravity as well, which is downwards, and that’s -9.81 m/s^2. After that,
all we need is 2 equations to find at what speed the bag strikes the ground. So what 2 equations
should we use? When we look at the information we have, we notice that for horizontal motion,
it starts at 0m, and the ending location is also 0m. So we can’t use an equation for
horizontal motion in this question. That means we need 2 equations in the vertical direction.
Let’s use this equation first. We know the initial velocity in the vertical direction,
and we know acceleration. Let’s substitute those in, and then write another equation.
We can also use this equation. Again, we can substitute what we know. Now,
we have 2 equations, with 2 unknowns so that means we can solve them which gives us our
answers. This question is also fully solved step by step, please check the description.
For almost every question, it is beneficial to draw a simple diagram,
and write down all the given values. That lets you determine which equations to use,
in which case. In addition to that, some questions will deal with trigonometric functions. You will
come across a few questions where you need to change it up to get an answer. The biggest one to
remember is that sin/cos is equal to tan. In those cases, it’s important to use identities and get
the equation in a way where you can plug it into your calculator. So when you simplify equations,
keep that in mind. Let’s look at one last example. In this question, we need to determine the muzzle
speed. Let’s first put down our coordinate system, and I will place it at A. Using this simplified
diagram, we can write down what we know. For the horizontal direction, the projectile starts at a
location of 0m, and the final location is 20 m. The horizontal component of velocity is Vcos30
degrees. Now, for the vertical direction, the projectile starts at 0m. The final location is 10m
– 1.8 m, so 8.2 m. That’s because our coordinate system is 1.8m above the ground, so we only
need the difference of height between where the projectile starts from, and when it ends up. The
vertical component of velocity is Vsin30 degrees. Let’s not forget gravity, which is -9.81 m/s^2.
Now, it’s time to write 2 equations. We can write one for horizontal motion,
and one for vertical motion. For horizontal motion, we can use this one:
Let’s substitute what we know into this equation. We know the final location is at 20 m,
the initial location is at 0m, and velocity is Vcos30 degrees. We can set this aside
and write another equation for vertical motion. For that, we will use this one.
For this one, we know the initial location is at 0m, final location is at 8.2 m, and velocity is
Vsin30 degrees, and gravity is -9.81 m/s^2. We now have 2 equations with 2 unknows, so we just
need to solve the 2. This gives us our answer, which is 27.96 m/s. This question is fully solved
in another video, please check the description. The questions you will face in this chapter will
require you to use 2 equations, and then solve for 2 variables. Once you write down what you know,
you need 2 equations. Most of the time, one will be for horizontal motion,
and one for vertical motion. Rarely, you will get a question that requires only
vertical motion equations. After you substitute, simply solve the 2 equations any way you like.
Many other projectile motion questions are fully solved in other videos, please check through the
available videos in the channel. If you have any questions, or ways to improve these videos,
please comment below. Thanks so much for watching and good luck with your studies.