What is the maximum height reached by a flying cow launched from a castle?

[AdamNailor]

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 realized 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 don't 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.

Thanks in advance.

 

[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

 

[AdamNailor]

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

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

 

[mgb_phys]

At the top of it's flight the vertical velocity is 0 ( should have been u_y
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.

 

[AdamNailor]

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?

 

[AdamNailor]

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.

 

[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 = u_y t₁ + 1/2 g t₁²

Coming down:
s+19.3 = u_y t₂ + 1/2 g t₂²
But it is stationary at the top of it's flight so u_y=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:
(u_y t₁ + 1/2 g t₁²) + 19.3 = 1/2 g (5.4-t₁)²

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