Cannon ball fired at castle - kinematics

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SUMMARY

The discussion centers on calculating the height at which a cannonball passes over a castle wall after being fired from a cannon. The cannonball, weighing 10.0 kg, is launched at an angle of 20 degrees with an initial speed of 100 m/s from a height of 1.0 m. The wall is located 500 m away and has a height of 24.3 m. The calculated height of the cannonball above the wall is 19.81 m, determined by separating the horizontal and vertical components of motion and applying kinematic equations.

PREREQUISITES
  • Understanding of kinematic equations, specifically Δx = Vit + 1/2at²
  • Knowledge of projectile motion, including horizontal and vertical components
  • Familiarity with trigonometric functions for resolving initial velocity
  • Basic grasp of gravitational acceleration (9.81 m/s²)
NEXT STEPS
  • Study the derivation and application of the kinematic equations in projectile motion
  • Learn how to resolve vectors into horizontal and vertical components
  • Explore the effects of different launch angles on projectile trajectories
  • Investigate the impact of air resistance on projectile motion calculations
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to enhance their teaching of kinematics concepts.

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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