Optimizing Cannon Range on a Vertical Tower

[jasonchiang97]

Homework Statement

A cannon that is capable of firing a shell at speed v₀ is mounted on a vertical tower of height h that overlooks a level plain below.

Show that the elevation angle α at which the cannon must be set to achieve maximum range is given by the expression

csc²(α) = 2(1+gh/V₀²)

Homework Equations

x(t) = v₀tcosα
y(t) = v₀tsinα - (1/2)gt² + h

The Attempt at a Solution

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First, I used

x(t) = v₀tcosα
y(t) = v₀tsinα - (1/2)gt² + h

I solved for t in y(t) and plugged it into x(t) to get

x = [v₀cosαsinα+v₀cosα(v₀²sin²α+2gh)^1/2]/g

then I solved for v₀cosα(v₀²sin²α+2gh)^1/2

and squared both sides to get

(xg - v₀²cosαsinα)/(v₀²cos²α) = v₀²+2gh

I expanded the left side of the equation and simplified with the right hand side to get

(g²/v₀²)x² -(2gtanα)x - 2gh = 0

Solved for x to get

x = v₀²/2(sinα + √(sin²α +2h/v₀²))

I took the dx/dα = 0 to get the maximum angle

dx/dα = 1+sinα(sin²α +2h/v₀²)^-1/2 = 0

And I said that it is max when sinα = 1 so then

x_max = v₀²/2 + v₀²/2(1+2h/v₀²

I'm unsure on how to obtain cscα from this

 

[RPinPA]

Your equations can't be what you said they are.

x(t) = v0tcosα
y(t) = v0tsinα

That says that both the horizontal and vertical motion are at constant velocity. The vertical motion is accelerated, as it is subject to gravity.

x = [v0cosαsinα+v0cosα(v02sin2α+2gh)1/2]/g

This looks like, contrary to the equations you said you were using, you used an equation that had a g in it, unlike the above. But I'm having a little trouble figuring out exactly what you're doing.

So what equations did you actually start with and what were your actual steps? Clearly they weren't what you typed here.

 

[jasonchiang97]

Your equations can't be what you said they are.
That says that both the horizontal and vertical motion are at constant velocity. The vertical motion is accelerated, as it is subject to gravity.This looks like, contrary to the equations you said you were using, you used an equation that had a g in it, unlike the above. But I'm having a little trouble figuring out exactly what you're doing.

So what equations did you actually start with and what were your actual steps? Clearly they weren't what you typed here.

Oops I meant to write y(t) = v₀tsinα - (1/2)gt² + y₀

where y₀ is just h, the height

so y(t) = v₀tsinα - (1/2)gt² + h

 

[RPinPA]

And then what did you do. It looks like you applied the quadratic formula to the equation for $y(t)$ to solve for t in terms of y. Is that so? Why not say so? Why not explain your calculations instead of leaving us to guess if you want us to follow and comment on what you're doing?

The remainder of your calculations seem to be following appropriate steps though I haven't tracked it in detail. Two comments on the final results:
1. $\csc(\alpha) = 1/\sin(\alpha),$ so if you have $\sin(\alpha)$ it's pretty easy to get $\csc(\alpha)$.
2. You say the maximum range is when $\sin(\alpha) = 1$, but that's when $\alpha$ is straight up and thus the range is 0. The cannonball lands on your head. Any angle less than that is going to give more range. So clearly there's an error somewhere.

3. There is something else wrong with your final steps. As I just said, if $\sin(\alpha) = 1$, then the cannon is pointing directly upward. Also $\cos(\alpha) = 0$ which means x(t) = 0 for all t, so I don't know what steps you followed to get a nonzero $x_{max}$ from that. But clearly you're doing something invalid.

I think I see the first place that seems to be going wrong anyway.

I took the dx/dα = 0 to get the maximum angle

That's an appropriate thing to do, though x may also have a minimum when $dx/d\alpha = 0$. So if you find more than one such point, you'll need to check which is which.

And I said that it is max when

Why are you looking for where $dx/d\alpha$ is maximized? You just said you wanted to know where $dx/d\alpha$ is 0.

 

[haruspex]

x = [v₀cosαsinα+v₀cosα(v₀²sin²α+2gh)^1/2]/g

Typo? The expression is dimensionally inconsistent. Need a ² on the first v₀.

squared both sides to get

(xg - v₀²cosαsinα)/(v₀²cos²α) = v₀²+2gh

The left hand side is dimensionless, the right is not.

dx/dα = 1+sinα(sin²α +2h/v₀²)^-1/2 = 0

Time for a sanity check. If you plug in h=0 do you get the standard result of 45 degrees?