# Person throws ball in elevator problem

Gold Member
Frankly, I am lost in all those terms. I would let vBH = u - (a + g)T, where T = t - t1, and integrate that. Then I would convert back to t and t1 if required.
So then x(t) - x(t1) = ut -ut1 - 1/2(a + g)T^2 = u(T) - 1/2 (a+g)T^2

What is the value of x at t1, by the way?
It is not given so instead let x(t) - x(t1) = Δx

It is not given so instead let x(t) - x(t1) = Δx

It is given. When somebody throws something, what is the relative distance at the time of throw?

Gold Member
It is given. When somebody throws something, what is the relative distance at the time of throw?
It is still with you so 0. So we do know x(t1) since we are working relative to the hand frame. We don't know x(t1) relative to the earth,right?

Correct, we do not know the distance relative to the Earth, nor the velocity, for that matter. But that is not in the question, either.

So how do you interpret the result physically?

Gold Member
Correct, we do not know the distance relative to the Earth, nor the velocity, for that matter. But that is not in the question, either.

So how do you interpret the result physically?

Relative to the hand, the ball has acceleration a+g, (i.e acceleration > g). This makes sense since the hand is obviously stationary wrt to itself. So the term comes from the upward acceleration of the lift and gravity.

Or do you mean that this eqn holds independent of where the ball is initally projected?

Well, physically the non-inertial frame manifests itself by gravity greater than usual. Apart from this all the other properties of motion are normal.

Gold Member
Well, physically the non-inertial frame manifests itself by gravity greater than usual.

Could you explain a bit more what you mean here?
Many thanks

What I mean is that the equation of "free fall" in the elevator frame is the same as in the Earth frame, except that the acceleration is g + a. So everything behaves as if the gravity were greater than "usual". Basically, you cannot say if it is the acceleration of the frame, or gravity that is not g. This will be important in your study of general relativity.

Gold Member
Many thanks for all your help today.

Going back to my first question I posed in the OP: Imagine an elevator going down at constant acceleration a and another going down at constant deceleration a.

I would have thought that the normal forces acting on someone inside the elevator be different (normal for acc > normal for dec). But how can I show this mathematically?

In both cases, the velocity of the lift is downwards. Take positive y downwards then ma = mg -N => N = m(g-a). What would change for the other case to yield a different normal?

The direction of motion (velocity) does not matter. The direction of acceleration does. As you say, the normal force is m(g - a). If a is positive (downward), then the force is smaller. If negative (upward), the force is greater. Which is in sync with intuition - my intuition, at least. I am not sure what else you would need to show.

Chestermiller
Mentor
vel. of ball relative to earth: u + vo. Do I have to consider the velocity of the hand? The problem statment says the hand stays fixed.

Mans acceleration rel. to earth: a
Balls acceleration rel. to earth = ##\frac{d}{dt} \left(u(t) + v_o \right)##, but I think the hand stays stationary, so acc. rel. to earth =## \dot{u}##

Yes. You have to consider the initial velocity of the hand relative to the earth (at time t = 0). Relative to the earth, the initial velocity of the hand is v0.

You got the man's acceleration relative to the earth correct. But, the ball's acceleration relative to the earth is minus g.

Are you familiar with the equation $x=x_0+v_0t+\frac{1}{2}at^2$?

Now all you have to do is write the corresponding equation for the location of the ball relative to the earth.

Gold Member
Yes. You have to consider the initial velocity of the hand relative to the earth (at time t = 0). Relative to the earth, the initial velocity of the hand is v0.

You got the man's acceleration relative to the earth correct. But, the ball's acceleration relative to the earth is minus g.

Yes, I see that now (I was confusing my frames when I wrote this, in particular u(t) as voko pointed out)- I think I have done all the steps in my previous posts with the aid of voko and my answer is 11 posts above.
Thanks

Chestermiller
Mentor
Yes. Sorry. I must have skipped over that previous post. Good job.

Chet