Man shot out of cannon question

  • Thread starter kdog3682
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In summary, to find the distance traveled, you can simply use the x-component of velocity, which is 30cos(36.86)*14 = 335.42 feet. To find the final velocity, you can use basic kinematic equations without needing to know the displacement, as you know the initial velocity, acceleration, and time of flight.
  • #1
kdog3682
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http://img263.imageshack.us/my.php?image=helpxy5.jpg
(pic attached)

Questions:
How far did man travel?
What was velocity upon hitting the wall? ( is it sufficent to cause death)

Known:
Time from cannon shot --> impact = 14 seconds ( on pic it says 16, but its actually 14)
Angle of cannon = 36.86 degrees
velocty = 30m/s

Homework Equations



vf^2 = vi^2+ viT ?

The Attempt at a Solution



i know that to find the distance traveled, is just the x-component.

THerefore distance = 30cos(36.86)*14 = 335.42 feet?

To find final velocity, I would think to use vf^2 = vi^2+ viT, but this is only for linear motion.

Please help!
 
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  • #2
kdog3682 said:
http://img263.imageshack.us/my.php?image=helpxy5.jpg
(pic attached)

Questions:
How far did man travel?
What was velocity upon hitting the wall? ( is it sufficent to cause death)

Known:
Time from cannon shot --> impact = 14 seconds ( on pic it says 16, but its actually 14)
Angle of cannon = 36.86 degrees
velocty = 30m/s

Homework Equations



vf^2 = vi^2+ viT ?

The Attempt at a Solution



i know that to find the distance traveled, is just the x-component.

THerefore distance = 30cos(36.86)*14 = 335.42 feet?

To find final velocity, I would think to use vf^2 = vi^2+ viT, but this is only for linear motion.

Please help!
when the man is going to land on the other building, he is going to land at angle thus he will have both X and Y components to his velocity. Since there is no force (thus no acceleration) in the X direction, the X component of his velocity will remain the same. Thus you must calculate the Y component of his velocity as well. Are the buildings the same height - thus what was his displacement in the Y direction? WHat is the acceleration i tneh Y direction only? What was his initial velocity? For how long was he in the air? Find the final velocity in the Y direction from this.

now you can calculate his final velocity using both X and Y components
 
  • #3
https://www.physicsforums.com/newreply.php?do=newreply&noquote=1&p=1345358[/b]

"Thus you must calculate the Y component of his velocity as well.
Are the buildings the same height - thus what was his displacement in the Y direction? WHat is the acceleration i tneh Y direction only? What was his initial velocity? For how long was he in the air? Find the final velocity in the Y direction from this.

now you can calculate his final velocity using both X and Y components"

You do not know if the buildings are the same heights. There is also the chance, that he lands on top of the building.
the Y displacement is 30sin(36.86)14 + 3 (the 3 is from the canon's height) - 4.9(14^2) = -693.88 feet.

A = -9.81?
Vi intial = 30 m/s or 0?
air time = 14s
using equation, vi^2 = 30^2 + 2(-9.81)*335 ( 335 is delta x) which equals
-5672.7... which is not the right answer. can sombody verify the right answers T_T
 
Last edited by a moderator:
  • #4
alright if you don't know the displacement, its fine

you still know the initial velocity, the acceleration, the time of flight and thus you can find the final velocity

you don't need to find the displacement to find the final velocity...
 

1. How does a man survive being shot out of a cannon?

The man is equipped with a special protective suit that absorbs the impact of being shot out of the cannon. The suit also has a built-in parachute that allows for a safe landing.

2. How do they aim the cannon to ensure the man's safety?

The cannon is carefully calibrated to a specific angle and direction, taking into account factors such as wind speed and distance. A team of trained professionals also monitors the shot and makes adjustments if necessary.

3. How fast does the man travel when being shot out of a cannon?

The speed can vary depending on the specific cannon and conditions, but on average, the man can reach speeds of around 60 miles per hour.

4. What training does a person need to be shot out of a cannon?

The person must undergo extensive training, including physical conditioning and learning how to properly position their body during the shot to minimize injury. They also receive instruction on how to safely land and exit the cannon.

5. Has anyone ever died from being shot out of a cannon?

Thankfully, there have been very few recorded deaths from being shot out of a cannon. This is due to the extensive safety measures and training that goes into each performance.

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