3-Dimensional Projectile Motion Homework 6. The object we are trying to hit has displacement vector...

3-Dimensional Projectile Motion

Homework 6

• The object we are trying to hit has displacement vector as follows-

d =(dx + vxt,dy + vyt,dz+vzt)( I’m going to use BIG V to represent the velocity of the projectile and little v to represent the velocity of what we are trying to hit)

• We are assuming we are launching our projectile from the origin.

• With that said our overall velocity vector needs to be broken up into x,y, and z directions for analysis.

• Vertical diagram

V z

VxVy

θ

ϕ

V

So then V = Vx2 +Vy

2 +Vz2

Vx =Vcosθ cosϕVy =Vcosθ sinϕVz =Vsinθ

X-Analysis

In any collision the projectile being launched must be at the same position after time t as the object we are trying to hit

dx + vxt=(Vcosθ cosϕ )t

This is the x position of the flying object at time t

This is the x position of the projectile at time t

Y- Analysis• The y analysis is much the same but just

involves the projectiles position on the y axis

dy + vyt=V(cosθ sinϕ )t

This is the y position of the flying object at time t

This is the y position of the projectile at time t

Z-Analysis

• The Z direction is different since gravity is accelerating the projectile at

−10m

s2

1

2at 2 +V(Sinθ)t+d0 =df

The projectile starts on the ground so d_0=0

−5t 2 +V (Sinθ )t = dz + vztThe projectiles position in z again has to equal that of the flying object in z

Flying object z position at time t

Projectile z position at time t

Combined Analysis

• We now need to combine the analysis from each direction to yield a system of equations- Lets remind ourselves what we already had-



−5t 2 +V (Sinθ )t = dz + vzt

Find Theta

• Okay so now we are going to find theta using our z analysis


V (Sinθ )t = dz + vzt + 5t 2

Sinθ =(d z+vzt + 5t 2 )

Vt

θ = Sin−1 (dz + vzt + 5t 2 )

Vt

⎛

⎝⎜⎞

⎠⎟

(Addition of 5t^2)

(Divide by V and t)

(Take the inverse sine)

Still finding theta

• Now we’ve got a formula for theta. Let’s see it again

θ =Sin−1 (dz + vzt + 5t 2 )

Vt

⎛

⎝⎜⎞

⎠⎟

We can then make a triangle with the following sides

θ

Vtdz + vzt+ 5t2

Still finding thetaCopying the Triangle on the previous slide

dz + vzt+ 5t2Vt

θ

We are going to use the Pythagorean theorem to get the other side like so

c2 =a2 +b2

c2 −b2 =a2

V2t2 −(dz + vzt+ 5t2 )2 =a

Still finding ThetaRedrawing our triangle again (this is hard work!)

dz + vzt+ 5t2Vt

θ

V 2t 2 −(dz + vzt+ 5t2 )2

Why would we do this? This allows us to know what cosine is

cosθ =V2t2 −(dz + vzt+ 5t2 )2

Vt

Now for phi!

• Okay lets file that away for a bit and think about phi. We had 2 equations involving phi:

dx + vxt=(Vcosθ cosϕ )tdy + vyt=V(cosθ sinϕ )t

Now were going to divide the one on the top by the one on the bottom!

dy + vyt=V(cosθ sinϕ )tdx + vxt=V(cosθ cosϕ )tdy + vytdx + vxt

=tanϕ

Phi-ing

Lets see that again: dy + vytdx + vxt

=tanϕWe now know what tan phi is- but both my equations involve sin and cos so I would really want to know what those are- to do this we draw yet another triangle

dy + vyt

dx + vxtϕ

Phi-phi-pho-fum• Redrawing our triangle

ϕdy + vyt

dx + vxt

We solve for the hypotenuse using the Pythagorean theorem again

c2 =a2 +b2

c2 = dx + vxt( )2+ dy + vyt( )

2

c= dx + vxt( )2+ dy + vyt( )

2

Phi-nal PhiRedrawing the triangle one more final time

ϕ dy + vyt

dx + vyt

c = dy + vyt( )2+ dx + vyt( )

2

We realize thatcosϕ =

dx + vxt

dx + vxt( )2+ dy + vyt( )

2

sinϕ =dy + vyt


2

Realizing what we know

We now know thatcosϕ =

dx + vxt


2

sinϕ =dy + vyt


2

cosθ =V2t2 −(dz + vzt+ 5t2 )2

Vt

As we have now defined the terms involving angles in our equations entirely with t we are ready to begin to solve by plugging in for each cos and sin.



Solvin’• We want to plug in the correct values for cos and sin here- if we read them

off the previous page and put them in we get


dx + vxt=VV2t2 −(dz + vzt+ 5t2 )2

Vt

⎛

⎝⎜⎜

⎞

⎠⎟⎟

dx + vxt


2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟t


2= V2t2 −(dz + vzt+ 5t2 )2


2=V2t2 −(dz + vzt+ 5t2 )2

dx2 + 2vxdxt+ vx

2t2 +dy2 + 2dyvyt+ v2

yt2 =V2t2 −(dz + vzt+ 5t2 )2

Where were we?

That last line was dx

2 + 2vxdxt+ vx2t2 +dy

2 + 2dyvyt+ v2yt

2 =V2t2 −(dz + vzt+ 5t2 )2

We want to expand the right side too so lets look at just the right side of the equation for the momentV 2t 2 −(dz + vzt+ 5t2 )2

V2t2 − dz + vzt+ 5t2( ) dz + vzt+ 5t2( )

V2t2 −(dz2 +dzvzt+ 5dzt

2 + vzdzt+ vz2t2 + 5vzt

3 + 5dzt2 + 5vzt

3 + 25t4 )

V2t2 −(dz2 + 2dzvzt+10dzt

2 + vz2t2 +10vzt

3 + 25t4 )

4th power term• So lets just look at this equations and collect the terms of power 4.

dx2 + 2vxdxt+ vx


yt2 Left side

V 2t 2 −(dz2 + 2dzvzt+10dzt

2 + vz2t2 +10vzt

3 + 25t4 ) Right side

Looks like there is only 1 4th power term: 25t 4

Okay so now lets look at third power terms

3rd power term• So lets just look at this equations and collect the terms of power 3.

dx2 + 2vxdxt+ vx


yt2 Left side


2 + vz2t2 +10vzt


Looks like there is only 1 3rd power term:10vzt3

Okay so now lets look at second power terms!

2nd power term• So lets just look at this equations and collect the terms of power 2.

dx2 + 2vxdxt+ vx


yt2 Left side


2 + vz2t2 +10vzt


Looks like there are 5 2nd power terms: vx

2t 22 + vy2t2 + vz

2t2 +10dzt2 −V2t2

Here I had to make sure the signs were respected- the stuff on the right changes signs as I bring it over to the left

1st power term• So lets just look at this equations and collect the terms of power 1.

dx2 + 2vxdxt+ vx


yt2 Left side


2 + vz2t2 +10vzt


Looks like there are 3 1st power terms: 2vxdxt + 2dyvyt+ 2dzvzt


0th power term• So lets just look at this equations and collect the terms of power 0.

dx2 + 2vxdxt+ vx


yt2 Left side


2 + vz2t2 +10vzt


Looks like there are 3 0st power terms: dx2 +dy

2 +dz2


Put it all together

• The whole equation then with all the terms moved over to the left side is

25t 4 +10vzt3 + vx

2t22 + vy2t2 + vz

2t2 +10dzt2 −V2t2 + 2vxdxt+ 2dyvyt+ 2dzvzt+dx

2 +dy2 +dz

2 =0

Factoring makes this look a lot nicer

25t 4 +10vzt3 + vx

2 + vy2 + vz

2 +10dz −V22( )t2 + 2vxdx + 2dyvy + 2dzvz( )t+dx2 +dy

2 +dz2 =0

if we call the initial velocity of the object we are trying to hit v and its net displacement D then we can write this equation like

25t 4 +10vzt3 + v2 +10dz −V22( )t2 + 2vxdx + 2dyvy + 2dzvz( )t+ D2 =0

25t4 +10vzt3 + v2 +10dz −V22( )t2 + 2 vv•

vd( )t+ D2 =0

So lets use this puppy now

• So I want to Solve HW problem 6 with our equation we have lovingly crafted

25t 4 +10vzt3 + vx

2 + vy2 + vz


2 +dz2 =0

In problem 6 v =(150,259.81,0)m/ s

d =(1000,2000,9000)m

V =500m/ sSo lets plug and chug:

25t 4 + 0t3 + (−69998.76)t2 + (1339240)t+ 86000000Put this in poly root finder, poly simult or graph it to find out…..

It ain’t going to happen Rick

• Solving for t reveals only imaginary and negative times could solve the problem- so our target is unhittable with our given technology.

Let’s do one that could work-

• Lets change homework 6 problem 6 tov =(40,70,0)m/ sd =(1000,2000,4000)m

V =500m/ sPlug and chug this into our equation:

25t 4 +10vzt3 + vx

2 + vy2 + vz


2 +dz2 =0

25t 4 + (−203500)t2 + 360000t+ 21000000 =0

Poly root finder tells us that t =88.72s or t=11.18s

Get the angles• So t was the following, and we can use this to get the angles by plugging

back into our original equations- lets use the first time first

t =88.72s or t=11.18s


−5(88.72)2 + 500(sinθ )(88.72) = 4000

500g88.72sinθ = 4000 + 5(88.72)2

sinθ =4000 + 5(88.72)2

500g88.72= .9774

θ = 77.79°

Now get phi

• We have θ =77.79°

Now we need phi= we use one of our original equations to find it

t =88.72sand

dx + vxt=(Vcosθ cosϕ )t1000 =500cos(77.79)cosϕ (88.72)

1000500g88.72gcos(77.79)

=cosϕ

ϕ =83.88°Great so if we fire the projectile at a vertical angle of 77.79 and a horizontal angle of 83.88 we will hit the target in 88.72s!

One more time though…• There is another possibility however- we could also hit the target in 11.18

sec. We do the same thing to find the angles this time


−5(11.18)2 + 500(sinθ )(11.18) = 4000

500g11.18sinθ = 4000 + 5(11.18)2

sinθ =4000 + 5(11.18)2

500g11.18= .8274

θ = 34.17°

And we find phi again

• So we find phi for this second time again

dx + vxt=(Vcosθ cosϕ )t1000 =500cos(34.17)cosϕ (11.18)

1000500g11.18gcos(34.17)

=cosϕ =.0205

ϕ =88.83°

Final Answers

• So we had two possibilities for theta, t and phiθ1 = 77.79°,ϕ 1 = 83.88°, t1 = 88.72s

θ2 = 34.17°,ϕ 2 = 88.83°, t2 = 11.18sWe should verify that all is well by checking the net speed of the projectile and making sure it is 500m/s- then we know theta and phi are matching up correctly and there wasn’t any mistakes

Double checking

We will just check for 88.72 sec only.In our x direction the target went a total

distance of d =vxtd=(40m/ s)(88.72s)d=3550m

It started however at 1000m away from our projectile, so to hit it our projectile would have to travel 4550m

d =Vxt4550m=Vx(88.72s)Vx =4550m/ 88.72sVx =51.28m/ s

Double Check y• Same thing for y- we know that the target started 2000m away from the

origin and continued at a speed of 70m/s. Its final position is

d =vyt

d=(70m/ s)(88.72s)d=6210.4m

We need to add 2000m to this since that’s where it started

So thend =Vyt

8210.4m=Vy(88.72s)

Vy =8210.4m/ 88.72s

Vy =92.54m/ s

Double Check z

To double check Z lets first find Vz with the Pythagorean theorm

So then 500 = Vx2 +Vy

2 +Vz2

500 = 2629.64 + 8563.65 +Vz2

250000 −2629.64 −8563.65 =Vz2

Vz =488.67m/ s

We verify with

Vz =Vsinθ =500sin 77.79°( )Vz =488.68m/ s

On the nose. So our formula works and we have a very powerful equation for handling almost any collision

Simple Archer Problem?

• We could even apply our equation to the simple archer problem.

25t 4 +10vzt3 + vx

2 + vy2 + vz


2 +dz2 =0

For the archer most of this stuff is 0, his target isnt moving and its at ground level so it only has displacement in 1 direction lets say x- then it all boils down to

25t 4 + −V22( )t2 +dx2 =0

and we also know that time to the target in x is

dx =Vcosθt

Arches away

• So we just had25t 4 + −V22( )t2 +dx

2 =0

Lets find t t 2 =V2 ± V4 −100d2

x

50dxVt

=cosθWe also had cosθ =

dx

V +V2 ± V4 −100d2

x

50

θ =cos−1dx

V +V2 ± V4 −100d2

x

50

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

Pluging them into each other yields

Not that complicated

• This looks complicated but with numbers its not

If an archer fires at a level target with V=20m/s that is dx=30m away, what angles can he fire at? First find t

25t 4 −400t2 + 900 =0Poly root finder tells us

t =3.65s or t=1.65s

Now just find theta

dx =Vcosθt

cosθ =dx

Vt=

3020g3.65

=.411

θ =65.73°or

cosθ =dx

Vt=

3020g1.65

=.909

θ =24.6365.73+ 24.63≅90

They are compliments so this seems kosher

3-Dimensional Projectile Motion Homework 6. The object we are trying to hit has displacement vector...

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