# Homework Help: Question about a flying cow

1. Nov 25, 2008

This is part of a project I'm doing, and I'm not sure if you can actually work out the answer with the information I have available, but I'll see. Its a review of the physics involved of the scene in the monty pythons the hold grail where the cow is catapulted out of the castle. So, the question would be something like this.

A cow is catapulted out of a castle and becomes visible at a height of 19.3 metres above ground level, and eventually hits the ground 5.04 seconds later. Find the maximum height reached by the cow, and the intial velocity the cow was launghed at at ground level inside the castle. g can be taken as 9.81, and assuming no air resistance.

My attempt: I used $$s=vt+\frac{1}{2}at^2$$ to give a height of 24.96 metres with v=0. But I then realised that this is not using the time from the launch, but only the time after this when the cow becomes visible above the castle wall. So I dont think this answer is correct. I'm not sure you work this out or not now. I think you need more info.

Are there any other ways I could estimate what it would be? Even if its not entirely accurate that would be better than nothing.

2. Nov 25, 2008

### mgb_phys

This is the same problem as the shooting a cannon upward from a cliff 19.3m high.
You have to do it in stages:
1, from 19.3m high upto the zenith, you know g,v but not u, T1
2, from top to the ground, you know g,u but not T2
You know T1+T2

3. Nov 25, 2008

But for number 2 you say I know what u is, but I dont, thats what I need to find out.

4. Nov 25, 2008

### mgb_phys

At the top of it's flight the vertical velocity is 0 ( should have been uy
You only need to worry about the vertical velocity to get the total height, we aren't concerned with how far it goes hiorizontaly.

Then everything else comes from s = ut + 1/2 at2 for each section.

5. Nov 26, 2008

ok so considering the motion after the peak we know that s=0.5*9.81*2.52^2 = 31.14m

That looks more like it. Thats correct yes?

6. Nov 26, 2008

Hang on, no that can t be right, because the 5.04 seconds is not the time taken for the whole trajectory, just for the bit where the cow becomes visible. So when I use 2.52 it actually should be a number a bit larger than that. Now I'm stuck.

7. Nov 26, 2008

### mgb_phys

Let s = the vertical distance from the top of the wall to the zenith, and t1 be the time.
Then s+19.3 is the vertical distance form the top of the flight to the ground, and t2 is the time.
Using s=ut+1/2at2

Going up:
s = uy t1 + 1/2 g t12

Coming down:
s+19.3 = uy t2 + 1/2 g t22
But it is stationary at the top of it's flight so uy=0 here.
so: s+19.3 = 1/2 g t22 and you know t1+t2=5.4

So you end up with the slightly ugly:
(uy t1 + 1/2 g t12) + 19.3 = 1/2 g (5.4-t1)2

The only thing left is the initial vertical velocity uy, but you know that:
v2 = u2 + 2gs and the final velocity (v) on the way up is 0.