Projectile motion of a cannon ball versus time

  • #1

Homework Statement


A cannon is inclined about the horizontal by an angle of 45° and it launches projectiles with initial velocity of ##v_p=300m/s##. A small airplane moves horizontally with a constant velocity of ##-100m/s##(so it is directed to the cannon) at high of ##500m##. If at ##t=0s##, the position of the airplane is ##7000m## away from the cannon, determine:

a) the time of which the projectile must be launch to hit the airplane.
b) what are the coordinates of the impact? at what time?


Homework Equations



The result of point a) is wrong, I don't understand why?

The Attempt at a Solution


starting with b)

From the system:
$$y=y_0+v_0sin(45°)-1/2gt^2$$
$$x=x_0+v_0cos(45°)$$

I found:
$$t=2.5s$$ and $$x=531m$$

so, the total time is $$t=\frac{x_(impact)-x_(airplane)}{v_(airplane)}64.7s$$

point a)

this should be ##64.7s-2.5s=62.2## because ##2.5s## is the time of the projectile to reach the coordinates ##(531m,500m)##.



Thanks for helping..
 
Last edited:

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


A cannon is inclined about the horizontal by an angle of 45° and it launches projectiles with initial velocity of ##v_p=300m/s##. A small airplane moves horizontally with a constant velocity of ##-100m/s##(so it is directed to the cannon) at high of ##500m##. If at ##t=0s##, the position of the airplane is ##7000m## away from the cannon, determine:

a) the time of which the projectile must be launch to hit the airplane.
b) what are the coordinates of the impact? at what time?


Homework Equations



The result of point a) is wrong, I don't understand why?

The Attempt at a Solution


starting with b)

From the system:
$$y=y_0+v_0sin(45°)-1/2gt^2$$
$$x=x_0+v_0cos(45°)$$

I found:
$$t=2.5s$$ and $$x=531m$$

so, the total time is $$t=\frac{x_(impact)-x_(airplane)}{v_(airplane)}64.7s$$

point a)

this should be ##64.7s-2.5s=62.2## because ##2.5s## is the time of the projectile to reach the coordinates ##(531m,500m)##.


Thanks for helping..
Let the cannon be located at (0,0) and the plane start from ##(-L,H) = (-7000,500)## at time ##t=0##. (Here I am supposing that distances and velocities to the right are positive and to the left are negative.) If the cannon fires to the left at time ##t_0##, what are the cannonball coordinates ##(x_c(t), y_c(t))## at a time ##t > t_0?## What are the airplane coordinates ##(x_p(t),y_p(t)## at time ##t > 0?## At the collision point/time you need ##(x_c(t),y_c(t)) = (x_p(t),y_p(t)),## and that gives two equations for the two unknowns ##t## and ##t_0.##

There are two roots for ##t_0## and ##t##; what are their physical meanings?

How would you do the question if the cannon fires to the right, so the cannonball chases a plane that has already past?
 
  • #3
tnich
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Homework Statement


A cannon is inclined about the horizontal by an angle of 45° and it launches projectiles with initial velocity of ##v_p=300m/s##. A small airplane moves horizontally with a constant velocity of ##-100m/s##(so it is directed to the cannon) at high of ##500m##. If at ##t=0s##, the position of the airplane is ##7000m## away from the cannon, determine:

a) the time of which the projectile must be launch to hit the airplane.
b) what are the coordinates of the impact? at what time?


Homework Equations



The result of point a) is wrong, I don't understand why?

The Attempt at a Solution


starting with b)

From the system:
$$y=y_0+v_0sin(45°)-1/2gt^2$$
$$x=x_0+v_0cos(45°)$$

I found:
$$t=2.5s$$ and $$x=531m$$

so, the total time is $$t=\frac{x_(impact)-x_(airplane)}{v_(airplane)}64.7s$$

point a)

this should be ##64.7s-2.5s=62.2## because ##2.5s## is the time of the projectile to reach the coordinates ##(531m,500m)##.



Thanks for helping..
I don't see anything wrong with your solution other than typos. Since this equation ##
y=y_0+v_0tsin(45°)-1/2gt^2
## is quadratic, there is another possible solution.
 
  • #4
Let the cannon be located at (0,0) and the plane start from ##(-L,H) = (-7000,500)## at time ##t=0##. (Here I am supposing that distances and velocities to the right are positive and to the left are negative.) If the cannon fires to the left at time ##t_0##, what are the cannonball coordinates ##(x_c(t), y_c(t))## at a time ##t > t_0?## What are the airplane coordinates ##(x_p(t),y_p(t)## at time ##t > 0?## At the collision point/time you need ##(x_c(t),y_c(t)) = (x_p(t),y_p(t)),## and that gives two equations for the two unknowns ##t## and ##t_0.##

There are two roots for ##t_0## and ##t##; what are their physical meanings?

How would you do the question if the cannon fires to the right, so the cannonball chases a plane that has already past?
Sorry if the problem was unclear but the planes starts at ##(L,H) = (7000,500)##

I don't see anything wrong with your solution other than typos. Since this equation ##
y=y_0+v_0tsin(45°)-1/2gt^2
## is quadratic, there is another possible solution.
Ok, so I guess the solutions were wrong.
I'd exclude ##t_2## since it's about ##40s##..
 
  • #5
Ray Vickson
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Sorry if the problem was unclear but the planes starts at ##(L,H) = (7000,500)##



Ok, so I guess the solutions were wrong.
I'd exclude ##t_2## since it's about ##40s##..
You were not unclear. I took the plane to be flying from west to east and the cannon pointing west. You are taking the plane flying from east to west and the cannon pointing east. Neither of us is wrong, but whatever convention you use ought to be spelled out explicitly.
 
  • #6
tnich
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Sorry if the problem was unclear but the planes starts at ##(L,H) = (7000,500)##



Ok, so I guess the solutions were wrong.
I'd exclude ##t_2## since it's about ##40s##..
I don't see anything wrong with 40.7s. Why would you exclude it?
 
  • #7
Ray Vickson
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Sorry if the problem was unclear but the planes starts at ##(L,H) = (7000,500)##



Ok, so I guess the solutions were wrong.
I'd exclude ##t_2## since it's about ##40s##..
There are two legitimate solutions. As I asked you in #2, can you explain the physical meanings of the two solutions? To enhance your understanding, try to give plots of the airplane and cannonball orbits in the two cases.
 
  • #8
There are two legitimate solutions. As I asked you in #2, can you explain the physical meanings of the two solutions? To enhance your understanding, try to give plots of the airplane and cannonball orbits in the two cases.
Well, I guess ##t_2## is the time it takes to reach the maximum heigh and come back to ##500m## before it drops into the terrain. Since the total time is about ##42.5s-40s=2.5s##.
 
  • #9
Ray Vickson
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Well, I guess ##t_2## is the time it takes to reach the maximum heigh and come back to ##500m## before it drops into the terrain. Since the total time is about ##42.5s-40s=2.5s##.
Right. So the cannonball can hit the plane while it (the cannonball) is rising, or it can hit the plane while it is falling. That is why I suggested you draw pictures.
 
  • #10
tnich
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Since the total time is about ##42.5s-40s=2.5s##.
Wait, the time for the cannonball to reach a height of 500m (the second time) is 40.7s, right? So what do 42.5s and 2.5s represent in this equation?
If the intersection of the trajectories is at 40.7s, what is the horizontal distance traveled by the cannonball at that point? How does that compare to the initial distance of the airplane from the cannon?
 
  • #11
tnich
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Ok, so I guess the solutions were wrong.
I'm not sure that is true. You haven't investigated all of the solutions, yet. I have found four potential solutions and you have only looked at one of them so far.
 
  • #12
I'm not sure that is true. You haven't investigated all of the solutions, yet. I have found four potential solutions and you have only looked at one of them so far.
Four? If I choose the second option, which is ##40s## the projectile will completely miss the airplane.. what are the other 2 solution?
 
  • #13
tnich
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Four? If I choose the second option, which is ##40s## the projectile will completely miss the airplane.. what are the other 2 solution?
The ones where the aircraft is flying west and the cannon is pointing west.
 
  • #14
PeterO
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I don't see anything wrong with 40.7s. Why would you exclude it?
In 40.7 seconds the cannonball will have travelled nearly 8500m "down range" - so unless the plane is moving away from the cannon it could never be there, no matter how fast it is flying.
 
  • #15
Thanks so much for the help and your time guys btw..
 
  • #16
tnich
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In 40.7 seconds the cannonball will have travelled nearly 8500m "down range" - so unless the plane is moving away from the cannon it could never be there, no matter how fast it is flying.
Once the plane flies over the cannon, it would be flying away from it. Before you completely reject this as a possible solution, consider that this kind of "tail chase" is a tactic used in real situations.
 

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