Cannon ball fired at castle - kinematics

AI Thread Summary
A cannon fires a 10.0 kg cannonball at a 20-degree angle with an initial speed of 100 m/s, aimed at a castle 500 m away with a wall height of 24.3 m. The approach involves separating horizontal and vertical motion components, using kinematic equations to calculate time and displacement. The time to reach the wall is determined to be approximately 5.321 seconds. The vertical displacement at that time is calculated to be 43.11 m. Ultimately, the cannonball passes 19.81 m above the castle wall.
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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