3-Dimensional Projectile Motion

1. 3-Dimensional Projectile Motion Homework 6

2. The object we are trying to hit has displacement vector as follows- ( 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)

3. 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.

4. Vertical diagram

5. 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 This is the x position of the flying object at time t This is the x position of the projectile at time t

6. Y- Analysis • The y analysis is much the same but just involves the projectiles position on the y axis This is the y position of the flying object at time t This is the y position of the projectile at time t

7. Z-Analysis • The Z direction is different since gravity is accelerating the projectile at The projectile starts on the ground so d_0=0 The projectiles position in z again has to equal that of the flying object in z Projectile z position at time t Flying object z position at time t

8. 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-

9. Find Theta • Okay so now we are going to find theta using our z analysis (Addition of 5t^2) (Divide by V and t) (Take the inverse sine)

10. Still finding theta • Now we’ve got a formula for theta. Let’s see it again We can then make a triangle with the following sides

11. Still finding theta Copying the Triangle on the previous slide We are going to use the Pythagorean theorem to get the other side like so

12. Still finding Theta Redrawing our triangle again (this is hard work!) Why would we do this? This allows us to know what cosine is

13. Now for phi! • Okay lets file that away for a bit and think about phi. We had 2 equations involving phi: Now were going to divide the one on the top by the one on the bottom!

14. Phi-ing Lets see that again: We 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

15. Phi-phi-pho-fum • Redrawing our triangle We solve for the hypotenuse using the Pythagorean theorem again

16. Phi-nal Phi Redrawing the triangle one more final time We realize that

17. Realizing what we know We now know that 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.

18. 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

19. Where were we? That last line was We want to expand the right side too so lets look at just the right side of the equation for the moment

20. 4th power term • So lets just look at this equations and collect the terms of power 4. Left side Right side Looks like there is only 1 4th power term: Okay so now lets look at third power terms

21. 3rd power term • So lets just look at this equations and collect the terms of power 3. Left side Right side Looks like there is only 1 3rd power term: Okay so now lets look at second power terms!

22. 2nd power term • So lets just look at this equations and collect the terms of power 2. Left side Right side Looks like there are 5 2nd power terms: 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

23. 1st power term • So lets just look at this equations and collect the terms of power 1. Left side Right side Looks like there are 3 1st power terms: 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

24. 0th power term • So lets just look at this equations and collect the terms of power 0. Left side Right side Looks like there are 3 0st power terms: 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

25. Put it all together • The whole equation then with all the terms moved over to the left side is Factoring makes this look a lot nicer 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

26. So lets use this puppy now • So I want to Solve HW problem 6 with our equation we have lovingly crafted In problem 6 So lets plug and chug: Put this in poly root finder, poly simult or graph it to find out…..

27. 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.

28. Let’s do one that could work- • Lets change homework 6 problem 6 to Plug and chug this into our equation: Poly root finder tells us that

29. 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

30. Now get phi • We have and Now we need phi= we use one of our original equations to find it 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!

31. 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

32. And we find phi again • So we find phi for this second time again

33. Final Answers • So we had two possibilities for theta, t and phi We 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

34. Double checking We will just check for 88.72 sec only. In our x direction the target went a total distance of It started however at 1000m away from our projectile, so to hit it our projectile would have to travel 4550m

35. 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 We need to add 2000m to this since that’s where it started So then

36. Double Check z To double check Z lets first find Vz with the Pythagorean theorm On the nose. So our formula works and we have a very powerful equation for handling almost any collision We verify with

37. Simple Archer Problem? • We could even apply our equation to the simple archer problem. 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 and we also know that time to the target in x is

38. Arches away • So we just had Lets find t We also had Pluging them into each other yields

39. 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 Poly root finder tells us

40. Now just find theta They are compliments so this seems kosher