How to Solve for missing mass in a Atwood's Machine?

[Not a Wrench]

Homework Statement

A gun is fired from a 200 meter cliff 200 meters away from an Atwood's Machine that is also suspended 200 meters in the air with a monkey of 20kg on the ground and a second unknown mass pulling the first mass up. Find the mass of the unknown needed for the bullet and the monkey to meet.

Homework Equations

a/2 * t = h - deltaY
T= 2*9.8/m2+20

I have no clue what else. I basically wrote everything that pertains to an atwood's machine

The Attempt at a Solution

I first found the time at which the bullet would travel the 200 meters, which is 4/15 seconds. I then tried to solve the acceleration. I get to a= 9.8(m2-20/m2+20). I don't know where to go from here. I have a T formula with m2 but the a I just got wouldn't help me plug into that. Help?

 

[TSny]

Hello. Did you state all of the information that was given? For example, do you know the direction that the bullet was fired? Does the Atwood's machine start from rest at the moment the bullet is fired?

 

[Not a Wrench]

Hello. Did you state all of the information that was given? For example, do you know the direction that the bullet was fired? Does the Atwood's machine start from rest at the moment the bullet is fired?

The gun is fired along horizontally along the x axis. The Atwood machine has a constant acceleration. I'm working of a drawing of a problem that was in class today. It has all the information needed to solve:http://imgur.com/y7D08C4
I have done everything possible for Atwood machines but I keep getting an acceleration in the 1000s m/s^2.

 

[haruspex]

The gun is fired along horizontally along the x axis.

So where exactly will the bullet be after 4/15s?

 

[Not a Wrench]

So where exactly will the bullet be after 4/15s?

It will hit the monkey at 199.65 meters vertically. So the monkey needs to travel 199.65 meters in 4/15 seconds. When I plug into the (a/2)4/15=199.65 I get an acceleration of 1118.05 m/s^2 which seems wrong. Then plugging into the a=g*(m2-m1/m2+m1) equation I get a mass of -20.3537 for the other box?! I have no clue what I am doing wrong here.

 

[haruspex]

It will hit the monkey at 199.65 meters vertically. So the monkey needs to travel 199.65 meters in 4/15 seconds. When I plug into the (a/2)4/15=199.65 I get an acceleration of 1118.05 m/s^2 which seems wrong. Then plugging into the a=g*(m2-m1/m2+m1) equation I get a mass of -20.3537 for the other box?! I have no clue what I am doing wrong here.

Hmmm.. the numbers in the question don't make any sense. There is no way the monkey could be raised far enough in the time.
The 750m/s is rather fast, but not crazily so. (About 2 to 6 times common values.) You'd have to slow it to about 45m/s to make it feasible.
Alternatively, is the 200m range wrong? It would work if that should be at least 3400m.

Please check very carefully that you have related the problem correctly.

 

[JBA]

Write out the standard equation for s in terms of a & t and compare that to your numerical equation you used to get your a value.