Cannonball Question Solved - 200m Target Hit?

  • Thread starter Thread starter juliany
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on a physics problem involving a 3-tonne cannon firing a 20kg cannonball horizontally at a speed of 250m/s towards a target 200 meters away, positioned 1 meter above ground level. The recoil velocity of the cannon is determined using the conservation of momentum principle. For the second part, the time for the cannonball to hit the target is calculated using kinematic equations, confirming that the cannonball will hit the target before striking the ground, assuming no air resistance.

PREREQUISITES
  • Understanding of conservation of momentum
  • Familiarity with kinematic equations
  • Basic knowledge of projectile motion
  • Concept of gravitational acceleration (g)
NEXT STEPS
  • Study the conservation of momentum in inelastic collisions
  • Learn how to apply kinematic equations for projectile motion
  • Explore the effects of air resistance on projectile trajectories
  • Investigate the calculation of recoil velocities in various scenarios
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding projectile motion and conservation laws in real-world applications.

juliany
Messages
11
Reaction score
0
Solved!

Homework Statement


A 3 tonne cannon fires a 20kg cannonball horizontally at a speed of 250m/s at a target 200 metres away. The cannon is 1 metre above ground level
a. Determine the recoil velocity of the cannon.
b. Assuming the cannon is wel aimed, will the cannon ball hit the target before it strikes the ground?


Homework Equations


a. I am unsure about this one, but would you determine the force of the cannon firing the cannonball and incorporate Newton's third law?

b. Would you determine the time for the cannonball to hit the target and the time for it to hit the ground? I am unsure which equation to use for this one as well:confused:



The Attempt at a Solution


Kind of answered this is relevant equations:blushing:
 
Last edited:
Physics news on Phys.org
The first part uses conservation of momentum. There are no forces involved because the velocity is constant.

Anything related to mass (including force) is unnecessary for the second part as it is just a kinematics problem. You're already given initial velocity v_{0x}=250\frac{m}{s} and v_{0y}=0\frac{m}{s}. If you're assuming there's no air resistance, then v_{0x} is constant so x(t)=v_{0x}t and y(t)=-\frac{g}{2}t^{2}+v_{0y}t+y_{0}.
 
zcd said:
The first part uses conservation of momentum. There are no forces involved because the velocity is constant.

Anything related to mass (including force) is unnecessary for the second part as it is just a kinematics problem. You're already given initial velocity v_{0x}=250\frac{m}{s} and v_{0y}=0\frac{m}{s}. If you're assuming there's no air resistance, then v_{0x} is constant so x(t)=v_{0x}t and y(t)=-\frac{g}{2}t^{2}+v_{0y}t+y_{0}.

Thanks:smile:
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K