Friday, October 1, 2021

Projectile Motion

Physics Index

Where are we going with this? The information on this page relates to the skills needed to investigate and evaluate the graphical and mathematical relationship (using either manual graphing or computers) of one-dimensional kinematic parameters (distance, displacement, speed, velocity, acceleration) with respect to an object's position, direction of motion, and time.

Projectile Motion
So where's this going? (See what I did there?)

A circus clown climbs into a cannon and is shot into the net across the ring. Into the net. Not into the crowd. Not onto the floor. Every time. Every. Single. Time.

Is it luck? No. It is physics.

Particularly, it is the physics of projectile motion.


Working with projectile motion requires working with vectors. Quick review…

Any vector quantity can be broken down into perpendicular component vectors.

The effects of the component vectors can be examined independently.

P.S. The components of a quantity will be the same quantity. A velocity vector can't be separated into anything but velocity vectors.

The perpendicular nature of the competent vectors creates a lot of very useful math relationships! (Hey! Do that thing where you make a nearly illegible, mostly incomprehensible graph!)

In the image above, `R is the original vector. `Rx is the component of  `R that is in the x direction. `Ry is the component of  `R that is in the y direction (perpendicular to the one in the x direction).

They "typical" triangle is labeled with a, b, c and θ

`R corresponds to c, `Ry corresponds to a,  and `Rx corresponds to b, 

In light of that, the following math relationships exist:

c2 = a2 + b2     <---- Pythagorean theorem

sinθ = a/c
cosθ = b/c
tanθ = a/b

Source, 2021-10-01

Okay… So… back to that clown and the cannon…

Boom! S/he flies out of the cannon with some initial velocity at some angle and comes down some distance away. Supposedly, there is a net at that exact distance to catch him/her.

That's the plan. Seems easy enough.

If the angle and initial velocity stays the same each time the clown will hit the net each time (unless they go to a different planet where acceleration due to gravity is different).

So… the goal is to reach a net that is some specific distance away. But they have that going-up and going-down thing going on… They could end up being 10 m in the air as they pass over the net! Meaning they would miss the net!

So… with avoiding that terrible outcome as the goal, let's look at the whole projectile motion calculation process step-by-step.

What the clown engineers do is work backwards from the distance to the net to the velocity and angle of the projectile-clown.

We, however, will go froward. The typical question might look like this:

How far horizontally from the origin will a [thing that could be a clown] travel if it is projected away from the origin with a velocity of something and at an angle of something with respect to the ground?

How far? That sounds like a distance/displacement!

The "how far" question is answered by finding the final distance. We have an equation for that!

df = di + vi•t + 1/2at2

BUT!!! Look at the question again. We need to find how far horizontally… 

Thus, we are ONLY interested in the component of the given velocity that is horizontal. We need to break the velocity up into horizontal (down range, along the ground, in the direction of x, etc.) and vertical (up, into the sky, in the direction of y, etc.)

Good thing we have that really nice triangle thing to work with! Let's re-draw it for clown cannons!

Another very fancy image.


 v is the velocity of the cannon (canon is the camera and established, authoritative facts)  

vx is the horizontal velocity. vy is the vertical velocity. θ is the angle.

Back to the question… Reworded, we are to find the horizontal displacement; find dx. And… We have an equation for that!

d= d+ vi•t + 1/2at2

Since we are measuring from the origin, then d is zero. Once the cannon fires, there is no other acceleration, so a is also zero.

That simplifies things. A lot. 

d= vi•t

Since we are looking for the HORIZONTAL displacement, we need to use the vx value.

d= vx•t

Well, well, well… how shall we find vx? Back to math!

sinθ = vy/v
cosθ = vx/v
tanθ = vy/vx

We have c and θ and we are looking for b. Which equation has those three variables?

cosθ = vx/v

Solve for what we need (vx).

v • cosθ = vx

Okay… so…  The horizontal displacement is THAT times however long it is in the air (t). 

d= vx•t

 So, we plug the calculated value from above in for vx. We're halfway there.

Finding out how long it is in the air (t) isn't hard. But… it's complicated.

First, we find out how long it takes to stop going up. Then, we find out how long it takes to fall from however high it managed to go (time to come down).

The total flight time (t) is found as

t = tup + tdown

That seems to make sense…

So, how long does it go up? It goes up until gravity stops it. Until its velocity in the y direction = zero.

The formula we need to use is the velocity formula:

vf = vi + at

where vf is when it stops (zero),  v is positive (going up). It is the y component of v so…
v • sinθ = vy

And a is negative (going down). Plugging in, we get

0 = vy + (-9.81)tup

-vy = (-9.81)tup

v/ 9.81 = tup

Alright… we have tup … what about tdown?

tdown  is how long it takes to fall from the highest point (where vy = 0) back to the ground. We can use the distance equation (above) with and 9.81  to find maximum dy .

dy = vytup + 1/2atup2

where a = -9.81 m/s/s.

(Sometimes, you'll be asked to find this, so… yeah… that's a thing.)

Now that we know how high it is, we can reverse the math and find tdown. But, since all of the numbers are the same (we are just flipping direction),  tdown comes out to be the same as tup.

Finding the total time turns out to be really easy!

t = tup + tdown


tup = tdown


t = 2•tup

Now, if we look back to the beginning of the problem, we have all we need to solve for the horizontal displacement!

Use the distance equation to find the horizontal displacement: 

d= vx•t

where vx was v • cosθ = vx and was 2•tup.

If so inclined, we could make a GIANT MESS out of the process by substituting everything in all at once. Look at the bottom of the page for that!

Back to the basics…

How about we break it down into steps without all of the discussion?


So, if you are given theta and the velocity, here's a checklist sort of process…

  • #1 Draw the diagram and label everything.
  • #2 Find the component velocities in the x and y directions:
v • cosθ = vx
v • sinθ = vy

  • #3 Find the time up using vy and acceleration due to gravity (probably 9.81 m/s/s)
tup = vy  / 9.81

  • #4 Find the total time where…
t = tup + tdown


tup = tdown


t = 2•tup

  • #5 Use the distance equation to find the horizontal displacement: 

d= vx•t

where vx was found in step 2 and t was found in step 4.

  • #6 BONUS: Find the maximum vertical displacement:

dy = vytup + 1/2atup2

where a = -9.81 m/s/s,  vy was found in step 2 and tup was found in step 3. 

GIANT MESS But actually COOL

d= v• t
dv • cosθ • t
dv • cosθ • 2 • tup
dv • cosθ • 2 • v / 9.81
dv • cosθ • 2 • v • sinθ / 9.81
d= 2 • v2cosθ • sinθ / 9.81
            wait for the magic!
d= 2 • v2 • 1/2 • sin2θ / 9.81
d=  v2 • sin2θ / 9.81

Say you wanted to find v for some distance x?

9.81 • d=  v2 • sin2θ

(9.81 • dx ) / sin2θ =  v2 

 √ (9.81 • dx ) / sin2θ   = v

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