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Projectile Motion in 3rd, finding out where 2 paths cross

  • Thread starter Entheogenx
  • Start date
  • #1

Homework Statement


2 projectiles, one at the origin the other at (0,100,0)m, have initial speeds of 35 m/s and 55 m/s respectively, and have direction cosine angles (45,60,60) and (60,135,60) respectively. Both are under the influence of gravity and experience a force of air resistance that has an acceleration of a = (-1, 0, -2.81) m/s^2. Assuming z>0, find:
a) The location both objects strike the ground
b) Where in space their paths cross, if they do.


Homework Equations


S = Si + Vit + 1/2at^2 (Position equation)
Vf= Vi + at
D= Vit + 1/2at^2


The Attempt at a Solution


The work here is quite lengthy and it took me a while to write it down on paper. I am not asking for direct help in solving both a) and b) for this problem (but if possible it will be greatly appreciated). I just have an algebraic question for part b of this problem.

How do i go about in finding where two objects in 3d space cross? I took the substitution approach by making Path 1's total position = Path 2's total position in order to solve for one of the times. I used the quadratic formula, solved a time 2 then plugged it back to get time 1. Time 2 = 1.7098s and Time 1= 1.914s is what i got for the two times. What does this tell me though? they never meet? Or do i have to do something additional here? Also i have 3 equations here X direction, Y direction, and Z direction. When i did substitution method I only used Y direction to get T1 and I subbed it into the X direction equation to get the T2 number. Do I have to also use the Z direction equation, or is it ok to only use two equations?

Your feedback here would be much appreciated. Again I am not asking for help on solving this problem. Instead I am asking general algebra help in what approach to take in order to find when the two projectiles meet.

Thanks
 

Answers and Replies

  • #2
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How do i go about in finding where two objects in 3d space cross? I took the substitution approach by making Path 1's total position = Path 2's total position in order to solve for one of the times. I used the quadratic formula, solved a time 2 then plugged it back to get time 1. Time 2 = 1.7098s and Time 1= 1.914s is what i got for the two times. What does this tell me though? they never meet? Or do i have to do something additional here? Also i have 3 equations here X direction, Y direction, and Z direction. When i did substitution method I only used Y direction to get T1 and I subbed it into the X direction equation to get the T2 number. Do I have to also use the Z direction equation, or is it ok to only use two equations?
You are being asked to "find where in space the paths cross." I would interpret that as, "find the point where the curves traced out by the projectiles intersect, even if they occupy that point at different times." So if you can solve for t1 and t2 such that,

[tex]
(x_1(t_1), y_1(t_1), z_1(t_1)) = (x_2(t_2), y_2(t_2), z_2(t_2))
[/tex]

Then this point (x, y, z) is your answer. If we want the projectiles to actually collide, then we also require that t1 = t2, but I don't think that that is what the question is asking.

Your approach so far seems perfect to me, so if your arithmetic is right you're almost done. What you need to do now is check whether the t1 and t2 you obtained give the same z-value for the two projectiles. If they do, then the paths do meet. If not, not.

From kind of a broad perspective, you can see how this works. Each path is a 1-D curve in space. If the motion was on a plane, the paths would be "guaranteed" to meet somewhere, unless they were just too far apart. But in 3-D space, the curves can just miss each other. Your work so far only gives you two components in common (x and y) for this reason; only two components are forced to intersect. If you are "lucky" the third component intersects, and you can plug in the times to find out if it does. But it is not guaranteed to, which is why you seemed to run out of equations to solve.

I'd be happy to check your arithmetic, if you can clarify something for me -- what do "direction cosines" mean, and why are there three? We only need two angles to specify an initial direction, right?...
 
  • #3
Overall I have 2 equations for each projectile, 4 in total. 1 equation which has the initial position, initial velocity, and acceleration moving upwards, then a second equation which starts at your max height and is dropping toward the ground so in that case the initial position is at your max, your velocity final from before is your velocity initial in the second equation, and acceleration moving downwards.

So with that into consideration, do i need to figure out a t1 for the rising portion of the projectile, then a t1 for the dropping portion? Since that is the complete path correct from the moment it goes up until it reach back to the ground. Same goes for the second projectile and its "t2". Well i tried finding a t2 for the dropping part and the quadratic formula was giving me negative times so I think I might have done something wrong or that an indication they don't cross each other in the dropping part?

Anyway, i subbed in like you said my t1 and t2 into the equations and the X&Y positions where the same, but like you said "if your lucky" the Z one would be the same as well. It wasn't the case though, they where off by a difference of 18 or so. So does that indicate that they don't cross?

Direction cosine angles mean each angle is adjacent to each axis. So for the first projectile at 35m/s it's xyz component would be broken into (35cos45, 35cos60, 35cos60) same goes for projectile 2. The cartesian plane is your Z as your vertical, y as your horizontal and x would be going from left to right intersecting x and y.
 

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