Projectile Motion fired from a cannon

[ItsImpulse]

Homework Statement

A projectile is fired from a cannon at an angle β with the horizontal at a velocity V. The cannon is situated inside a cave. The cannon is situated at a distance X away from the inner most part of the cave. The cave makes an angle α with the horizontal. What is the angle β, in terms of V,α and X, will the cannon have maximum range? Assume no air resistance

Homework Equations

Kinematics Equations

The Attempt at a Solution

I have tried to create a function that has β as a function of V,α and X, however, when i try to equate the maximum height of the projectile to the height of the cave where the projectile is at maximum(I.E (x+0.5range)tan(α))= (0.5range)tanβ), my α is always affected by β. Any help?

 

[voko]

It is unclear: do we have situation 1 or situation 2?

 

[ItsImpulse]

It is situation 2, and the angle between cannon and horizontal is B

 

[voko]

Do I understand correctly that you assume that the highest point of the trajectory is given by $(R/2) \tan \beta$, where $R$ is the range for the given angle and initial velocity? I do not think this is correct.

 

[ItsImpulse]

Yes I did that assumption due to the fact that it is constant acceleration thus at half the range, the projectile will be at the maximum height of its trajectory. Is this wrong?

 

[voko]

Your equation seems to assume that the trajectory will be a straight line at angle $\beta$ with the horizontal. But can that be true? Does such a trajectory even have a highest point?

 

[ItsImpulse]

How would I go about solving it then? I don't know how else to attempt the question

 

[voko]

In the relevant equations part of the form that you filled out, you just wrote "Kinematics Equations". It is hard to tell what you know and what you don't. What is the shape of the cannon's ball trajectory? What is its maximum height?

 

[ItsImpulse]

I would think that the cannonball's trajectory would be a parabolic one. I don't know how to approach the problem as I normally deal with A level-standard questions so that's roughly where my knowledge is at.

 

[voko]

Yes indeed, the trajectory is a parabola.

Do you understand the concept of vectors and their components? Is motion accelerated in the horizontal component? In vertical component?

 

[ItsImpulse]

Yes I do. In the y- direction acceleration is -g if upwards positive. X direction has no acceleration as no air resistance.

 

[HallsofIvy]

How you handle this depends upon the formulas you have been given. The most basic is to say that the acceleration vector is <0, -g> where the first component is horizontal and the second component vertical. The velocity vector, after time t, is then &lt;Vcos(\beta), -gt+ Vsin(\beta)&gt; and so position, after time t, is &lt;Vcos(\beta)t, -(g/2)t^2+ Vsin(\beta)t&gt;, relative to the cannon's position. The "range" is the value of Vcos(\beta)t at the positive value of t that makes -(g/2)t^2+ V sin(\beta)t= 0. That happens when t= (2/g)V sin(\beta) and so the range is (2/g)V^2sin(\beta)cos(\beta).

The point of the cave is that we can never have the vertical height of the cannon ball, -(g/2)t^2+ Vsin(\beta)t, greater than the height of the cave, (Vcos(\beta)t)tan(\alpha).

 

[voko]

Then you should be able to write the equations for both components of displacement.

Edit: which HallsofIvy just did for you.

 

[voko]

The point of the cave is that we can never have the vertical height of the cannon ball, -(g/2)t^2+ Vsin(\beta)t, greater than the height of the cave, (Vcos(\beta)t)tan(\alpha).

I think that assumes that $X = 0$.

 

[ItsImpulse]

Assuming that x is not 0. I get that sinB - cosBtanA < (2xtanA-gt^2)/(2vt). I still cannot get B as a function of A, X and V?

 

[voko]

I believe you have a sign error in your inequality. Check it.

Apart from this, you should analyse the problem in two ways.

Forget that the cannon is in the cave. Just find the maximal range - at what value of $\beta$ does it happen? Then see what condition you have on $V$, $X$ and $\alpha$ so that the cannon can shoot at this optimal angle.

The second analysis pertains to the case when the above condition for optimal shooting does not hold. In that case the trajectory should just touch the cave at some point to attain maximum range. See what condition that gives you.

 

[ItsImpulse]

The maximum range would be at 45 degrees wouldn't it? The problem lies that the height of the cave is a function of B ,(ie. XtanA + tanAVcosBt) so how then can I solve explicitly B as a function of X,v,A?

 

[voko]

45 degrees is correct. What conditions must $V$, $X$ and $\alpha$ satisfy so that $\beta = 45 ^ \circ$ is possible?

 

[ItsImpulse]

Oh I see that's how you go about doing it. Would be accurate to say that the max range is at 45 degrees but with the cave inhibiting motion, it's the greatest angle possible?

 

[voko]

As I said earlier, there are two situations.

One is when the projectile motion is not restricted by the cave. Example: the wall of the cave is very far away or is almost vertical, or the initial velocity is very small. Then the max range is given by $\beta = 45 ^ \circ$. In this case $\beta$ cannot be expressed in terms of $V$, $X$ and $\alpha$ - it is simply $45 ^ \circ$. But in this case you need to find the mathematical formulation of "the projectile motion is not restricted by the cave".

The other situation is when this mathematical formulation is violated and "the projectile motion is restricted by the cave".