Cannonball Question Solved - 200m Target Hit?

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The discussion focuses on solving a physics problem involving a cannon firing a cannonball. The recoil velocity of the cannon can be determined using conservation of momentum principles. For the cannonball's trajectory, the time to hit the target and the time to hit the ground are calculated using kinematic equations, assuming no air resistance. The initial horizontal velocity is constant, while the vertical motion is influenced by gravity. Ultimately, the analysis confirms whether the cannonball will hit the target before striking the ground.
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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:
 
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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:
 
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