How Far Will a Human Cannonball Travel When Their Acceleration is 0?

[jean28]

Homework Statement

A man acts as as a human cannonball. When the cannon is fired, he has a velocity of 18ft/s. How far will the man have traveled when his acceleration is 0? When will this occur? Neglect friction.

Also given is:

Man is 120 lb-m

Homework Equations

kinematics equations:
v = v0 + a*t
x - x0 = Vo*t + 1/2 a*t^2
v^2 = Vo^2 + 2a (x - X0)
x - x0 = 1/2 (v0 + v) t
x - x0 = v*t - 1/2 a*t^2

The Attempt at a Solution

Im thinking about maybe using one of the equations without t to find the final x, and then use the final x to find t. Is that correct? Any help?

 

[Villyer]

Under what conditions is acceleration 0?

 

[jean28]

Under what conditions is acceleration 0?

There is a hint given: The acceleration is NOT constant (so this makes the equations irrelevant), and the moment when the acceleration is 0 is still INSIDE the cannon, not after he is shot.

 

[jean28]

Wait... am I missing something here? If it is asking how much the guy has moved INSIDE the cannon when the acceleration is 0 then doesn't that mean he hasn't moved at all?

 

[cepheid]

The moment of 0 acceleration does not occur inside the cannon. The question is asking how far he'll travel after being shot, before his acceleration becomes 0. You need to determine when that occurs. Hint: what forces act on him during flight?

 

[jean28]

The moment of 0 acceleration does not occur inside the cannon. The question is asking how far he'll travel after being shot, before his acceleration becomes 0. You need to determine when that occurs. Hint: what forces act on him during flight?

No, actually, I asked my professor and he just told me it was a trick question and he wanted to know INSIDE the cannon, not after the guy is shot from the cannon and the acceleration is 0.

 

[cepheid]

There is a difference between a trick question and an ambiguously-worded question.

You know the man's final kinetic energy upon release from the cannon, which means you know how much total work is done on him while inside of it.

I assume that the acceleration is 0 at the moment of release from the barrel, since that is the instant at which he first breaks contact with whatever was accelerating him.

That's all I've got.