Cannon on a vertical tower

  • #1

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

xmax = v02/2 + v02/2(1+2h/v02

I'm unsure on how to obtain cscα from this

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Last edited:

Answers and Replies

  • #2
RPinPA
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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.
 
  • #3
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
 
  • #4
RPinPA
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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.
 
Last edited:
  • #5
haruspex
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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?
 

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