**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.*

Boom!

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:

c<---- Pythagorean theorem^{2}= a^{2}+ b^{2}sinθ = a/ccosθ = b/ctanθ = 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 ofsomethingand at an angle ofsomethingwith 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 + 1/2at^{2}

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. |

So…

vis the velocity of the cannon(canon is the camera and established, authoritative facts)vxis the horizontal velocity.vyis the vertical velocity.θis the angle.

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

df = di + vi•t + 1/2at^{2}

Since we are measuring from the origin, then

*di*is zero. Once the cannon fires, there is no other acceleration, so*a*is also zero.That simplifies things. A lot.

df = vi•t

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

*vx*value.df = vx•t

*Well, well, well…*how shall we find

*vx*? Back to math!

sinθ =vy/vcosθ =vx/vtanθ =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*).dx = 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),*vi*is positive (going up). It is the y component of v so…v • sinθ =vy

And

*a*is negative (going down). Plugging in, we get0 = vy + (-9.81)tup-vy = (-9.81)tupvy / 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 = vy•tup+ 1/2atup^{2}

wherea = -9.81m/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 + tdownandtup = tdownsot = 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:

dx = vx•twherevxwasv • cosθ =vxandtwas2•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?*

_____________

SUMMARY

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 =vxv • 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 + tdownandtup = tdownsot = 2•tup

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

dx = vx•twherevxwas found in step 2 andtwas found in step 4.

- #6 BONUS: Find the maximum vertical displacement:

dy = vy•tup+ 1/2atup^{2}wherea = -9.81m/s/s,vywas found in step 2 andtupwas found in step 3.

_____________

GIANT MESS But actually COOL

*dx = vx • t*

*dx =*

*v • cosθ*

*• t*

*dx =*

*v • cosθ*

*•*

*2 • tup*

*dx =*

*v • cosθ*

*•*

*2 •*

*vy*

*/ 9.81*

*dx =*

*v • cosθ*

*•*

*2 •*

*v • sinθ*

*/ 9.81*

*dx = 2 •*

*v*

^{2}• cosθ*• sinθ*

*/ 9.81*

*wait for the magic!*

*dx = 2 •*

*v*

^{2}• 1/2 • sin2*θ*

*/ 9.81*

*dx =*

*v*

^{2}• sin2*θ*

*/ 9.81*

Say you wanted to find v for some distance x?

*9.81 • dx =*

*v*

^{2}• sin2*θ*

*(9.81 • dx ) /*

*sin2*

*θ*

*=*

*v*

^{2}

*√ (9.81 • dx ) / sin2θ = v*

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