Determining the starting position when dealing with an inclined launch

[arhzz]

Homework Statement

The cannonballs have a speed of V0 = 100m/s when they exit the barrel.The sparrows sit in a nearby castle. The castle stands on a 250m high mountain (= H = level of the castle courtyard), has a 37m (= h) height of the castle wall with a thickness d = 10m.
Note: Start your calculation with x (t) and z (t).
a) At which point x0 in front of the castle does he have to set up his cannon when the inclination angle α0 of the barrel is fixed at 60∘ and he wants to touch the castle battlement in the ascending branch of the parabola?

b) Which is the heighest point of the trajectory (zmax)

c) And at which xmax do the cannoballs hit the courtyard

Relevant Equations

t = vo * sinα/g

Now I did what the suggestion said I started my calculation with x(t);

x(t) = v0 * cosα * t

As I was missing the t component I found it like this

t = v0 * sinα/g

t = 8,82 s

Now I've put t back into x(t) and got this result;

x(t) = 441 m

Now what I think I have gotten is the complete flight path (or length) of the cannoball in the x coordinate.

Now how do I get the x0 component out of this? The way I am looking at this is that the y component (or z in my case) doesn't have to do anything with this,meaning I can leave that out. The thickness of the wall shouldn't play a roll in this either but the part that confuses me is the height.Since the castle is on the hill can I sum up the two heights and just get one and look at it that way. Also we are given a sketch that says that the 0,0 point in the coordinate system is the castle it self, meaning that if we were to put the cannon infront of the castle,we would have to go left out the 0,0 point meaning the result would be negative? The main question is how do I calculate the x0 position of the cannon?

 

[kuruman]

The time you calculated, 8.82 s, is the time required to reach maximum height. Since we want the cannonball to touch the castle battlement in the ascending branch of the parabola, this happens at a time before 8.82 s. This time could be relevant to part (b).

I suggest that you find an expression for the height of the cannonball above ground as a function of horizontal distance from the wall, $y(x)=\dots$ and then solve for the value of $x$ that corresponds to the height of the castle battlement. The $y(x)$ expression will also be useful for part (c).

On edit: Vertical axis $y$ here is misidentified. It should be $z$ as defined in the statement of the problem. See post #29.

 

[arhzz]

The time you calculated, 8.82 s, is the time required to reach maximum height. Since we want the cannonball to touch the castle battlement in the ascending branch of the parabola, this happens at a time before 8.82 s. This time is relevant to part (b).

I suggest that you find an expression for the height of the cannonball above ground as a function of horizontal distance from the wall, $y(x)=\dots$ and then solve for the value of $x$ that corresponds to the height of the castle battlement. The $y(x)$ expression will also be useful for part (c).

Yes that with the time I have figured out, I've actually solved the b part of the question.

Now this part is a litle fuzzy to me "Find an expression for the height of the cannonball above ground as a function of a horizontal distance from the wall". What exactly does that mean. I should figure out the height of the cannonball at what point? I don't really understand what is meant by this.

 

[kuruman]

You have two kinematic equations, one, $x(t)$ gives you the horizontal position as a function of time and the other, $y(t)$ gives you the vertical position as a function of time. Use the first equation to find an expression for $t$ in terms of $x$ and put that in the second equation to find $y$ in terms of $x$. This will give you the height above ground as a function of the horizontal distance. No specific point is involved; if you have a specific value of $y$ or $x$, you can plug it in the equation to find the corresponding specific value of $x$ or $y$. See how it works?

 

[arhzz]

You have two kinematic equations, one, $x(t)$ gives you the horizontal position as a function of time and the other, $y(t)$ gives you the vertical position as a function of time. Use the first equation to find an expression for $t$ in terms of $x$ and put that in the second equation to find $y$ in terms of $x$. This will give you the height above ground as a function of the horizontal distance. No specific point is involved; if you have a specific value of $y$ or $x$, you can plug it in the equation to find the corresponding specific value of $x$ or $y$. See how it works?

Ahhh now I see what you mean, but that is really a math thing not a physics. Okay I'll try that and we will see what happens

 

[kuruman]

Ahhh now I see what you mean, but that is really a math thing not a physics. Okay I'll try that and we will see what happens

Math is the language of physics. Doing physics without math is like cooking without heat.

 

[arhzz]

Okay so I've done it like this

x(t) = v0 * cos α * t --> t = x/v0cos α
y(t) = v0 * sin α * t - 1/2 gt^2

Now I've replaced the t in y with the t I've gotten from x

y(t) = v0 * sin α * x/v0cos α - 1/2 g* (x/v0cos α )^2
y(t) = v0 * sin α * x/v0cos α - 1/2 g* x^2/cos^2 α

Now have I done this correctly? And if so what is my next step? putting in the values and trying to find x?

 

[kuruman]

Your next step is to find the value of $x$ for which $y$ has the desired value. What is that "desired value"? It would be easier for me and others to trouble shoot your expression if you substituted the value for α only and find the trig functions. Leave v0 and g as symbols until the very end.

 

[arhzz]

Your next step is to find the value of $x$ for which $y$ has the desired value. What is that "desired value"? It would be easier for me and others to trouble shoot your expression if you substituted the value for α only and calculate the trig functions. Leave v0 and g as symbols until the very end.

Okay, I'll do that

 

[kuruman]

Use $\sin(60^o)=\dfrac{\sqrt{3}}{2}~;~~\cos(60^o)=\dfrac{1}{2}$ and simplify.

 

[kuruman]

Yes, it's the square of the cosine itself, not the angle. What you have on the left hand side is not y(t) but y(x). You eleiminated time, remember? Also replace the values for the sine and cosine as I indicated in post #10 and post the equation that you get.

 

[arhzz]

Yes, it's the square of the cosine itself, not the angle. What you have on the left hand side is not y(t) but y(x). You eleiminated time, remember? Also replace the values for the sine and cosine as I indicated in post #10 and post the equation that you get.

Ah yes we eliminated time right.

y(x) = v0 * √3/2 * x/v0* 1/2 - 1/2 * g * x^2/v0 cos^2 *1/2

 

[kuruman]

There are simplifications to be made. The expression cos^2 *1/2 does not make sense. If the cosine is 1/2 what is it's square? You can do that in your head.

Also, if you are planning on posting here, I would suggest that you use LaTeX for writing equations. That will make your posts easier to read and you will find that people will be more eager to help you if they understand your writings better. Click on the link "LaTeX Guide" (lower left) to see how to do it. It's easy to learn and there are examples of everything you might need.

 

[arhzz]

There are simplifications to be made. The expression cos^2 *1/2 does not make sense. If the cosine is 1/2 what is it's square? You can do that in your head.

Also, if you are planning on posting here, I would suggest that you use LaTeX for writing equations. That will make your posts easier to read and you will find that people will be more eager to help you if they understand your writings better. Click on the link "LaTeX Guide" (lower left) to see how to do it. It's easy to learn and there are examples of everything you might need.

Your right, I'll look into LaTeX and post my equation again.

 

[arhzz]

$y(x) = v0 *\dfrac{\sqrt{3}}{2} * \frac {x} {v0 * \frac 1 2} - \frac 1 2 * g * \frac {x^2} {v0 * \frac 1 4}$

This is what I've gotten by simplifying, and LaTeX is pretty cool.

 

[archaic]

$y(x) = v0 *\dfrac{\sqrt{3}}{2} * \frac {x} {v0 * \frac 1 2} - \frac 1 2 * g * \frac {x^2} {v0 * \frac 1 4}$

This is what I've gotten by simplifying, and LaTeX is pretty cool.

you can use \times for a nicer looking multiplication sign. ;)

 

[arhzz]

you can use \times for a nicer looking multiplication sign. ;)

Oh must have over seen that, thanks!

 

[kuruman]

Look for simplifications to make your equation manipulation and eventual calculations easier. For example, in $y(x) = v0 *\dfrac{\sqrt{3}}{2} * \frac {x} {v0 * \frac 1 2} - \frac 1 2 * g * \frac {x^2} {v0 * \frac 1 4}$ you have $v0$ in the numerator and $v0$ in the denominator that cancel. Do you see other such cancellations?

 

[arhzz]

Look for simplifications to make your equation manipulation and eventual calculations easier. For example, in $y(x) = v0 *\dfrac{\sqrt{3}}{2} * \frac {x} {v0 * \frac 1 2} - \frac 1 2 * g * \frac {x^2} {v0 * \frac 1 4}$ you have $v0$ in the numerator and $v0$ in the denominator that cancel. Do you see other such cancellations?

I think I made a mistake with v0 it should be v0 squared, the one in the last fraction where x is squared;

Yes I've aware I could make these a bit easier by cancelling each other out, I'll try to implement that and we will see what happens

 

[arhzz]

$y(x) =\dfrac{\sqrt{3}}{2} * \frac {x} { \frac 1 2} - \frac 1 2 * g * \frac {x^2} {v0^2 * \frac 1 4}$Now I think we can cancel the $\frac 1 2$ and the $\frac 1 4$

$y(x) =\dfrac{\sqrt{3}}{2} * \frac {x} { \frac 1 2} - g * \frac {x^2} {v0^2 * \frac 1 2}$

Did I do this wrong and am I missing something.

 

[kuruman]

$y(x) =\dfrac{\sqrt{3}}{2} * \frac {x} { \frac 1 2} - \frac 1 2 * g * \frac {x^2} {v0^2 * \frac 1 4}$Now I think we can cancel the $\frac 1 2$ and the $\frac 1 4$

$y(x) =\dfrac{\sqrt{3}}{2} * \frac {x} { \frac 1 2} - g * \frac {x^2} {v0^2 * \frac 1 2}$

Did I do this wrong and am I missing something.

You are almost there. Can you get rid of the annoying $\frac{1}{2}$ fractions in the denominators? Also drop the asterisks and use implied multiplication. It's neater.

 

[arhzz]

You are almost there. Can you get rid of the annoying $\frac{1}{2}$ fractions in the denominators? Also drop the asterisks and use implied multiplication. It's neater.

Your answer didn't pop up on my feed. I think I have gotten the full simplification now

$y(x) = \sqrt 3 * x - 2 * g * \frac {x^2} v0^2$

 

[kuruman]

Small typo, it should be $y(x) = \sqrt {3 }x - \dfrac {2g} {v_0^2}x^2.$

Now what do you think you should do with it? If you forgot, read the first sentence in post #8.

 

[arhzz]

Small typo, it should be $y(x) = \sqrt {3 }x - \dfrac {2g} {v_0^2}x^2.$

Now what do you think you should do with it? If you forgot, read the first sentence in post #8.

Now I need to find the value of x for which y has the desired value. That means I need to find the desired value of y and try to find an x for it?

 

[kuruman]

Now I need to find the value of x for which y has the desired value. That means I need to find the desired value of y and try to find an x for it?

Yes.

 

[arhzz]

Okay so I think I've gotten the trick. The desired value of y(x) is actually the value of the height, so

y(x) = H + h meaning;

y(x) = 287

So now we have this equation;

$287 = (x) = \sqrt {3 }x - \dfrac {2g} {v_0^2}x^2.$

 

[kuruman]

Yes, if you solve it for $x$, you will get the values at which the height of the cannonball above ground is 287 m. I don't know what $(x)$ means, so I am ignoring it because it doesn't belong.

 

[arhzz]

Yes, if you solve it for $x$, you will get the values at which the height of the cannonball above ground is 287 m. I don't know what $(x)$ means, so I am ignoring it because it doesn't belong.

That also was a typo, but the values I get for x, are those the values for my x0? And also how do I solve this, using the p,q quadratic equation would be my first guess

 

[kuruman]

That also was a typo, but the values I get for x, are those the values for my x0? And also how do I solve this, using the p,q quadratic equation would be my first guess

How do you define x0? Variable $x$ stands for horizontal distance of the cannonball when it is vertical distance $y$ above ground. When you solve the quadratic, what do you get?

I don't know what the p,q quadratic equation refers to. I know that if you have the quadratic equation$ax^2+bx+c=0$, the solutions are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

On edit: I just noticed that the vertical axis is defined by the problem as $z$ whereas I used $y$ from the start. Let's agree to use $y$ for consistency and to avoid confusing anyone else who peruses this thread.

 

[arhzz]

How do you define x0? Variable $x$ stands for horizontal distance of the cannonball when it is vertical distance $y$ above ground. When you solve the quadratic, what do you get?

I don't know what the p,q quadratic equation refers to. I know that if you have the quadratic equation$ax^2+bx+c=0$, the solutions are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

On edit: I just noticed that the vertical axis is defined by the problem as $z$ whereas I used $y$ from the start. Let's agree to use $y$ for consistency and to avoid confusing anyone else who peruses this thread.

I have solved this issue with the quadratic formula and gotten that the x0 should be -221,15.Now to get xmax
And again your response did not pop up in my inbox