How Does Velocity Relate to Kinetic Energy in Projectile Motion?

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SUMMARY

The discussion centers on the calculation of the speed of a cannonball with a mass of 150 kg, which is fired with a kinetic energy of 447.9 kJ at an angle of 25.95 degrees. The initial calculation using the formula KE=1/2(mv^2) yielded a speed of 77.28 m/s, which was initially marked incorrect due to a typographical error in the mass, which should have been 15 kg. Participants emphasized the importance of significant figures and calculator mode (degrees vs. radians) when calculating the initial vertical velocity. Ultimately, the instructor confirmed the correct mass, validating the initial calculations.

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Stevemotto
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Hey guys, this question has boggled my mind not because I don't know how to approach it but apparently the method I'm using is incorrect!

Question
A cannonball with a mass of 150 kg is fired from a cannon. The cannonball leaves the cannon with 447.9 kJ of kinetic energy. The cannon makes an angle of θ = 25.950 o with the ground as shown in the diagram (not to scale). The castle wall is d = 3820 m from the cannon. Ignore any effects due to air resistance. The acceleration due to gravity is 9.80 ms-2. This castle is built in the future with special building materials that allow it to have very high walls.

What is the speed of the cannonball as it leaves the cannon?My Attempt

So considering I'm given KE & the mass and we want to find the velocity I used the formula KE=1/2(mv^2). I convert kJ to J as required and get 447900 = 1/2(150*v^2), solving for v; v=77.28 m/s.

I enter my solution and apparently it's wrong...

Can anyone evaluate my method and tell me if the system is crazy or I'm just approaching the question incorrectly. I have attached the image to the online quiz showing the question & diagram as well as the system saying I got it wrong when entering the supposed solution.

Kind Regards,

Steven
 

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I agree with your answer but not the number of significant figures.
IME with those online quizzes is that they only accept an answer with the correct sig. figs.
 
Actually you would think that, but the instructor has said that the no. of sig figs don't matter and that we should put as many as possible to avoid rounding off error.
 
I guess you've tried entering the answer to different decimal places?
 
Yes, to no avail..
 
Stevemotto said:
Yes, to no avail..
Did you try 77.3?
 
Yep, still incorrect. I'm having some thoughts of the interaction of gravity on the velocity but that doesn't make sense and we aren't given time. Thinking of emailing the instructor..
 
If you have trials left over, you could guess the 150 kg is a typo and should have been 15.0 kg ... :rolleyes:
 
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Your solution is correct given the numbers. I suggest that you show your solution to your instructor and ask him/her to explain why it was marked incorrect. Sometimes it happens that the solution is keyed in incorrectly. I think this is one of these times.
 
  • #10
kuruman said:
Your solution is correct given the numbers. I suggest that you show your solution to your instructor and ask him/her to explain why it was marked incorrect. Sometimes it happens that the solution is keyed in incorrectly. I think this is one of these times.
From the text it is obvious that the cannonball can't leave the cannon at 77 m/s. With the given angle it would only travel a puny fraction of the 3820 m mentioned in the problem statement. A cannonball of 150 kg is utterly unlikely. 15 kg gives a reasonable exercise, but I think the composer did a bad job by having to introduce very big walls so people won't be uncomfortable in a subsequent part of the exercise.

@Stevemotto , can you post the remainder of the exercise already, or do you first need to get past this initial hurdle ?
 
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  • #11
LOL, the instructor emailed back and said I was correct. The figure for the question was wrong as it should have been 15kg, BvU you predicted correctly :D
 
  • #12
Alright, this will be the last question I ask; so just a follow up on the next part of the question, they are us to calculate:

What is the initial vertical velocity of the cannon ball?

So I use V(y)=v*sin(theta)

From last part v=244.3768 & theta=25.95

Yielding, V(y)= 244.3768*sin(25.95)

Now the question is whether if I use radians or degree mode on my calculator as they give 2 different answers. I feel like it should be degrees as the question gave it degrees.
 
  • #13
Stevemotto said:
Now the question is whether if I use radians or degree mode on my calculator as they give 2 different answers. I feel like it should be degrees as the question gave it degrees.
If the argument of the sine function is specified in degrees, then use degree mode on your calculator.
 
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  • #14
Always check with an estimate: sin(approximately 30 degrees) = approximately 0.5 so if you get 0.4376 you're probably OK
 
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  • #15
Thanks guys, got it out!
 
  • #16
So why the remark on the future castle with ridiculously high walls ?
 
  • #17
We aren't given the height of the wall, 3820m is the distance from the wall to the cannon. But I was suppose to find the speed & mangnitude of the ball hitting the wall
 
  • #18
magnitude of a 15 kg cannonball :smile: ?
 
  • #19
yep!
 
  • #20
as in diameter in centimeters, given that it's made of iron ?
 
  • #21
What diameter? I got the magnitude by multiplying the mass of the cannon ball with the speed in which it hit the wall.
 
  • #22
Ah, the magnitude of the momentum of the cannonball. (the words 'of the momentum' fell off the panel :smile:).

You don't have to calculate how high on the wall this momentum does its damaging work ?
 
  • #23
No question on momentum :)
 

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