Particle shot from a height h on Earth

[Dustgil]

Homework Statement

A cannon that is capable of firing a shell at speed [tex]V_0[/tex] is mounted on a vertical tower of height h that overlooks a level plain below.

(a) Show that the elevation angle [tex]\alpha[/tex] at which the cannon must be set to achieve maximum range is given by the expression

[tex]csc^2(\alpha)=2(1+\frac {gh} {V_0^2})[/tex]

(b) What is the maximum range R of the cannon?

Homework Equations

[tex]F=m\frac {d^2r} {dt^2}[/tex]

The Attempt at a Solution

The only force on the particle is -mgk, so

[tex]m\frac{d^2r} {dt^2} = -mgk[/tex]

[tex]\frac{d^2r} {dt^2} = -gk[/tex]

[tex]\vec{v}=-gtk+v_0[/tex]
[tex]\vec{r}=\frac {-gt} {2}k+v_0t+r_0[/tex]

The initial velocity and position can be expressed as such

[tex]V_0=V_0cos(\alpha)i+V_0sin(\alpha)k[/tex]
[tex]r_0=hk[/tex]

Therefore vector r is

[tex]\vec{r}=V_0tcos(\alpha)i+(h+V_0tsin(\alpha)-\frac{gt^2} {2})k[/tex]

In component form,

[tex]x=V_0tcos(\alpha)[/tex]
[tex]z=(h+V_0tsin(\alpha)-\frac {gt^2} {2})k[/tex]

when z=0 (taking just the postitive root of the quadratic formula because we are concerned about positive t)

[tex]t= \frac {V_0sin(\alpha) + \sqrt{V_0^2sin^2(\alpha)+2gh}} {g}[/tex]

Plugging this into x, we get

[tex]x=V_0cos(\alpha)(\frac {V_0sin(\alpha)+\sqrt{V_0^2sin^2(\alpha)+2gh}} {g})[/tex]

This is the equation of motion in the x direction. It seems to make sense because if the angle is 90 degrees, we get 0 distance traveled. When the angle is 0 degrees, the y component of the initial velocity cancels out to give something nice. My thinking from this point is that taking dx/da and setting it equal to 0 would give the angle that would make x a maximum. However, when I do this I get a horribly complicated equation that I can't figure out how to simplify. Am I on the right track?

 

[mfb]

[tex]\vec{v}=-gtk+v_0[/tex]
[tex]\vec{r}=\frac {-gt} {2}k+v_0t+r_0[/tex]

Check the second equation. You got it right later so I guess it is just a typo.

The last equation looks good. The value of ##\alpha## maximizing this also corresponds to a value of ##\cos(\alpha)## maximizing this (and due to the range of ##\alpha## you don't have to deal with ambiguities like ##\pm 2 \pi##), so you can introduce a new variable ##p=\cos(\alpha)## and maximize x with respect to this one. That should make the expressions easier. Choosing ##p=\sin(\alpha)## could work as well.

You know the right answer already. You do not have to solve the equation, you just have to show that the answer works.

 

[Dustgil]

So

[tex]\frac {dp} {d\alpha} = -sin\alpha[/tex]

That would be correct right? I don't think I've done something like this exactly before. So if I plugged it in and got something like

[tex]\frac {dp} {d\alpha}p[/tex]

That would equal one right? Sorry if that's a dumb question.

 

[mfb]

That would be correct right?

Yes, but it does not matter. If you substitute before taking the derivative with respect to p ##\alpha## does not appear any more.

-sin(α)cos(α) is not equal to 1.

 

[SammyS]

Homework Statement

A cannon that is capable of firing a shell at speed ##\ V_0 \ ## is mounted on a vertical tower of height ##\ h\ ## that overlooks a level plain below.

(a) Show that the elevation angle ##\ \alpha \ ## at which the cannon must be set to achieve maximum range is given by the expression
[tex]\csc^2(\alpha)=2\left(1+\frac {gh} {V_0^2}\right)[/tex]
(b) What is the maximum range R of the cannon?
...

The Attempt at a Solution

...
Plugging this into x, we get
[tex]x=V_0\cos(\alpha)\left(\frac {V_0\sin(\alpha)+\sqrt{V_0^2\sin^2(\alpha)+2gh}} {g}\right)[/tex]
This is the equation of motion in the x direction. It seems to make sense because if the angle is 90 degrees, we get 0 distance traveled. When the angle is 0 degrees, the y component of the initial velocity cancels out to give something nice. My thinking from this point is that taking dx/da and setting it equal to 0 would give the angle that would make x a maximum. However, when I do this I get a horribly complicated equation that I can't figure out how to simplify. Am I on the right track?

Assuming the last equation is correct, the following may help.

Solve for ##\displaystyle \ \sqrt{V_0^2\sin^2(\alpha)+2gh} \,.\ ##

Then square both sides and do a little cancelling/simplifying.

Use implicit differentiation & set ##\ dx / d\alpha =0 \ . ##

Added in Edit:

Thanks mfb (for the "like"), but it didn't work out as well as I had hoped. Maybe I just didn't find a good way to eliminate x in the result .

You can get essentially the same thing my suggestion leads to, by completing the square in the expression for time, z = 0, then substituting the expression for x as a function of t. (Probably should be labeled t₀ for the time to reach zero elevation.)