# Mechanics - Castle under siege

1. May 3, 2006

### Natasha1

Need to do this by tomorrow guys just need some help with it. Could anyone help to get me started please :-)

A castle is under siege. Its walls are 40m high, and it stands on level ground. The besiegers have a cannon at ground level 200m away and aim directly towards the castle and an angle of 35 degrees to the horizontal.

Using the standard equation for path, or working from first principles, find the speed of firing for a cannonball to just clear the top of the wall of the castle and find the time taken?

The defenders also have a cannon placed at the top of the walls, and can only fire this horizontally. Their aim is to either hit the opposing cannon, or to hit the members of the opposing army strung out behind for a distance of a further 200m. What range of values of speeds of firing is appropriate for the defendersâ€™ cannonball?

PS: Air resistance is to be ignored in this question

2. May 3, 2006

Do you know "the standard equation of path" as mentioned in the question? If you do, then I think this problem can be easily solved.

3. May 3, 2006

### Natasha1

Is it y = x * tan (alpha) - (g/2u^2) sec^2 (alpha) * x^2

4. May 3, 2006

### GregA

$$y = xtan\theta - \frac {gx^2}{2u^2cos^2\theta}$$ ?Looks good to me...Don't know why but I can never fully remember that formula...always have to derive it or look it up but any way...
The only variable in there that you don't know is u ..with this you can then find t.

Last edited: May 3, 2006
5. May 4, 2006

### Natasha1

Have I got this right please.... could someone take 5 mins to check all calculations to see if I'm correct, and tell me if I have gone wrong somewhere. Thanks :-)

Right...

Solving forces horizontally:

constant speed of: u.cos alpha

so
x = u.cos alpha.t
200=u.cos 35. t (1)

Solving vertically:

a=-g
Initial velocity: u.sin alpha
y = u.t + 1/2 a.t^2
y = u. sin alpha . t - (1/2) g.t^2
40 = u. sin 35.t - (1/2) (10). t^2 (2)

From (1) we get u = 200 / cos 35.t

replacing in (2) we get:

40 = (200 / cos 35.t) sin 35.t - (1/2) g.t^2
40 = 200.tan 35 - (1/2) g.t^2
(1/2)g.t^2 = 200 tan 35 - 40
5t^2 = 200 tan 35 - 40
t^2 = 40 tan 35 - 8
t = 4.473 seconds

As
u = 200 / cos 35.t
u = 200 / cos 35 * 4.473
u = 54.584 ms-1

I actually found first t then u, is this still ok?

Last edited: May 4, 2006
6. May 5, 2006

### GregA

Can see no problems with your working or answer. Which you find first, v or t, doesn't really matter because both can be expressed in terms of the other and then substitutions made.

The only thing I would say is that you give your final values with accuracy to 3 d.p. given that you use g = 10 in addition to other assumptions this seems a bit odd. I would just say u =55 m/s and t = 4.5 secs

Last edited: May 5, 2006