Determining the starting position when dealing with an inclined launch

[arhzz]

That's fine. Your answer is correct in that case. The wall to cannon distance, a positive number, ought to be 221 m. I missed the point that the problem expects you to place the origin of coordinates at the base of the exterior wall. Onto xmax as you said.

I don't know why. I too sometimes see those and sometimes do not. There could be delays in the notification system. If I want to know whether there are new responses, I click on the bell and the new message notifications (if any) pop up regradless of whether a number is next to the bell. If you are concerned about this, send a private message to @Greg Bernhardt and ask about it.

Its fine,I'll just double check.So you said in a previous post that y(x) will be useful for xmax as well.Now to be straight with xmax, that should be the maximum distance of the cannon where it hits the wall right?

 

[kuruman]

Maybe this is relevant. You will find it under My PF/Preferences

 

[arhzz]

Maybe this is relevant. You will find it under My PF/Preferences

View attachment 271149

Yea that was already enabled, but as I said no big deal I'll just double check always.Back to the task at hand, xmax

 

[kuruman]

Its fine,I'll just double check.So you said in a previous post that y(x) will be useful for xmax as well.Now to be straight with xmax, that should be the maximum distance of the cannon where it hits the wall right?

The problem asks "at which xmax do the cannoballs hit the courtyard". In other words, how far away from the wall can the cannon be placed and still be able to land cannonballs in the courtyard without hitting the wall. Here is where the thickness of the wall becomes relevant.

BTW, what is your answer to part (b)?

 

[arhzz]

The problem asks "at which xmax do the cannoballs hit the courtyard". In other words, how far away from the wall can the cannon be placed and still be able to land cannonballs in the courtyard without hitting the wall. Here is where the thickness of the wall becomes relevant.

BTW, what is your answer to part (b)?

Okay so if the thickness of the wall becomes relevant that means that we need to consider it as another distance,thus adding it to the already existing one? For the b part my answer was

y(x) =382,26 m

 

[kuruman]

Your maximum height in (b) is correct. As a point of clarification and to get you to develop good habits, $y(x)$ is shorthand notation and a reserved generic name standing for "the height above ground at any distance $x$ from the wall." It cannot be used to mean anything else. The 382.26 m is the maximum height above ground at a specific distance from the wall, therefore it needs a specific name such as $y_{max}$ standing for "the maximum height above ground."

As for part (c), think of it this way. In part (a) you found how close the cannon can be and clear the wall. Now imagine moving the cannon back from that position. You need to find how far away the cannon can be and still clear the wall. What part of the wall will be hit first as the cannon is moved farther back?

 

[arhzz]

Your maximum height in (b) is correct. As a point of clarification and to get you to develop good habits, $y(x)$ is shorthand notation and a reserved generic name standing for "the height above ground at any distance $x$ from the wall." It cannot be used to mean anything else. The 382.26 m is the maximum height above ground at a specific distance from the wall, therefore it needs a specific name such as $y_{max}$ standing for "the maximum height above ground."

As for part (c), think of it this way. In part (a) you found how close the cannon can be and clear the wall. Now imagine moving the cannon back from that position. You need to find how far away the cannon can be and still clear the wall. What part of the wall will be hit first as the cannon is moved farther back?

Well if we move the cannon back the first part of the wall that is going to be hit, is the top of the wall

 

[kuruman]

Which part of the top of the wall? It is 10 m thick.

 

[arhzz]

I've solved it 480m is xmax

 

[kuruman]

I've solved it 480m is xmax

That is significantly less than what I got. If you substitute your answer in the equation for the height, you get $$h=\sqrt {3 }\times 480~\mathrm{m} - \dfrac {2\times 9.8~\mathrm{m/s^2}} {(100~\mathrm{m/s})^2}(480~\mathrm{m})^2=380~\mathrm{m}$$ which is way above the top of the wall. This says that the cannon can be moved farther back and still clear the wall. Hint: When you solved the quadratic in part (a), you got two solutions. What is the meaning of the second solution?

 

[arhzz]

That is significantly less than what I got. If you substitute your answer in the equation for the height, you get $$h=\sqrt {3 }\times 480~\mathrm{m} - \dfrac {2\times 9.8~\mathrm{m/s^2}} {(100~\mathrm{m/s})^2}(480~\mathrm{m})^2=380~\mathrm{m}$$ which is way above the top of the wall. This says that the cannon can be moved farther back and still clear the wall. Hint: When you solved the quadratic in part (a), you got two solutions. What is the meaning of the second solution?

Umm,okay well my teacher actually graded my paper and 480 was right.

 

[haruspex]

Umm,okay well my teacher actually graded my paper and 480 was right.

I'm not sure what part c is asking. I think it is asking for how far away from the outer wall can the cannon be placed and still have the cannonballs clearing the wall.
In solving the quadratic for part a, the √(b²-4ac) part is near enough √(3/4), which is half the magnitude of b. So if the smaller solution is 221 the larger is three times that. But that is a distance from the inner side of the wall, so the distance from the outer wall is 3x221-10=653m.

What interpretation are you using to get 480m?

 

[kuruman]

Umm,okay well my teacher actually graded my paper and 480 was right.

You may have gotten credit for 480 m from your professor, but this doesn't make the answer correct. Here is why. When you solve the quadratic $287=\sqrt{3}x-\frac{2g}{100^2}x^2$, you get $x_1=221~\mathrm{m}$ and $x_2=663~\mathrm{m}$. They are the two distances from the cannon at which the cannonball is 287 m vertically above the cannon. At 221 m, when the ball is ascending, it just grazes the upper outside corner of the wall;. At 663 m when the ball is descending, it will also hit the outside edge (we assume a point ball here) unless you move the canon closer by 10 m in which case the ball will graze the inside corner.

Answer: x_max = 653 m from the outer face of the wall.

Do with this information what you wish. As I was writing this, I noted the post by @haruspex and we agree.

 

[arhzz]

You may have gotten credit for 480 m from your professor, but this doesn't make the answer correct. Here is why. When you solve the quadratic $287=\sqrt{3}x-\frac{2g}{100^2}x^2$, you get $x_1=221~\mathrm{m}$ and $x_2=663~\mathrm{m}$. They are the two distances from the cannon at which the cannonball is 287 m vertically above the cannon. At 221 m, when the ball is ascending, it just grazes the upper outside corner of the wall;. At 663 m when the ball is descending, it will also hit the outside edge (we assume a point ball here) unless you move the canon closer by 10 m in which case the ball will graze the inside corner.

Answer: x_max = 653 m from the outer face of the wall.

Do with this information what you wish. As I was writing this, I noted the post by @haruspex and we agree.

I'll calculate it all again, I am just wondering if my teacher was wrong but thank you for you help (both of you) !

 

[kuruman]

I'll calculate it all again, I am just wondering if my teacher was wrong but thank you for you help (both of you) !

It could be that your teacher decided to give you full credit because your method was correct but the numerical answer was off for some silly reason like forgetting to square a quantity. That is why it is important to show all your work in as much detail as you can.

 

[arhzz]

It could be that your teacher decided to give you full credit because your method was correct but the numerical answer was off for some silly reason like forgetting to square a quantity. That is why it is important to show all your work in as much detail as you can.

Very much possible, again thank you for your help!