Two cannons, different heights and initial velocities

AI Thread Summary
The discussion revolves around the collision dynamics of two horizontally firing cannons at different heights and velocities. It explores whether the projectiles can collide if fired simultaneously and if delaying the shot from cannon B can ensure a collision with cannon A's projectile. The analysis reveals that the vertical motion equations indicate the projectiles cannot collide due to their differing initial heights. A mathematical approach is suggested to analyze the time and distance relationships necessary for a potential collision. Ultimately, the conclusion is that the projectiles cannot meet at the same height, confirming they cannot collide.
Walter303
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Homework Statement


Two cannons(A at inital height h; B at initial heigt h/2) fire(horizontally), B firing with higher velocity than A.

Homework Equations


1) If they fire at the same time, is it possible that the projectiles collide?
2) Is it possible for B to delay his shot so that his projectile always collides with A's?

The Attempt at a Solution


1) Their x and y coordinates must be the same in case of collision.
xA(t) = xB(t)

x0 + v0A*t = x0 + u*v0A*t
("u" is a positive constant. B's velocity is greater than A's)
v0A = v0B
(but v0A and v0B are different)
I'm not sure about what I should do next...2) I mean, I can get a certain example, but I can't seem to find any formal explanation. For example,
v0A = 10 m/s
v0B = 20 m/s
If B delays the shot by 1 s(tA = 2 s):
v0A*tA = v0B*tB
10*2 = 20*1
then it's possible
 
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Walter303 said:
1) Their x and y coordinates must be the same in case of collision.

Correct. , however if you can prove that the two y co-ordinates can't be the same then you have proved they cannot collide (without needing to think about the x co-ordinates at all). What does the motion in y look like?
 
CWatters said:
Correct. , however if you can prove that the two y co-ordinates can't be the same then you have proved they cannot collide (without needing to think about the x co-ordinates at all). What does the motion in y look like?

They both fire horizontally, so the initial velocity(in y) for both is zero:

h - (1/2)g*t^2 = h/2 - (1/2)*g*t^2

h = h/2

So, it seems that they can't collide. Still, I know there's something wrong here, I just don't know what...
 
This is a variation of the "monkey and the hunter" problem.
If A were aiming at B and B firing level then both bullets would fall the same amount away from the line of fire.
Now, for the second question try
1/2 g tx^2 + g tx * tB + 1/2 g tB^2 = h/2 + 1/2 g tB^2 where tx is the time that bullet A had been falling before bullet B was fired.
This requires both bullets to fall to the same level for collision to occur.
The times involved depend on the separation of the cannons and the speeds of the bullets.
 
" A were aiming at B and B firing level" - I think this should be if "A were aiming at B and B aiming at A"
Then both bullets would fall the same distance from the "line of fire" in time t and would end at the same
height at time t (whatever that is)
 
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