Cannon ball fired at castle - kinematics

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Homework Help Overview

The problem involves kinematics, specifically the motion of a cannonball fired at an angle towards a castle wall. The scenario includes calculating the height of the cannonball as it passes a wall that is 500m away and 24.3m high, given its initial speed and launch angle.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss separating horizontal and vertical components of motion, and the application of kinematic equations. There is an initial attempt to calculate the time to reach the wall and the subsequent vertical displacement.

Discussion Status

Some participants have provided guidance on using separate equations for horizontal and vertical motion. One participant has shared their calculations and results, indicating a progression in the discussion, though no consensus on the final answer has been reached.

Contextual Notes

There is mention of uncertainty regarding the role of mass in the calculations, and the original poster expresses confusion about including it in their approach.

yaser1989
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Homework Statement


A cannon fires a 10.0kg cannon ball 1.0m above the ground and at 20o to the horizontal with an initial speed of 100m/s. A castle 500m away has a wall 24.3m high. At what distance above the castle wall does the cannon ball pass?


Homework Equations


Δx = Vit + 1/2at2
It's the only one I can think of that applies.

The Attempt at a Solution


I thought of getting time it takes to get to the wall then using that to calculate displacement above the wall but I got the wrong answer. The only thing I don't know is how to include the mass (10.0kg) in my calculations.
 
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Anyone?
 


You want to separate the horizontal and vertical component of motion in this problem, which means that you separate the horizontal and vertical component of initial velocity and the horizontal and vertical motion equations. You already listed one motion equation, there should be another one for another direction. It's convenient if you let x be the horizontal component and y be the vertical component.
 


I got it.

I rearranged Δx = Vit + 1/2at2 for time:

t = Δx / Vi --> since acc'n in the x direction is 0, 1/2at2 = 0.

So then, t = 500/ 100cosΘ where Θ = 20o
= 5.321.

Then I used y = (Vi )yt + 1/2at2
= (100sinΘ)(5.321) + 1/2(-9.81)(5.3212 )
= 43.11

Subtracted that from the height of the wall - the height of the cannon:

43.11 - 23.3 = 19.81

Thanks for the help !
 

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