# Person throws ball in elevator problem

1. Feb 20, 2013

### CAF123

1. The problem statement, all variables and given/known data
An elevator moves up with a constant acceleration a and a person stands inside the elevator. At time $t_1$ > 0, the person throws a ball directly upwards with velocity u wrt himself. Find the distance of the ball from the man's hand. (assume he does not move his hand)

3. The attempt at a solution
I have the solution to this in my notes, but I tried it differently and I get to a step that I can't solve. I want to know if it is correct up to where I can't solve the resulting diff eqn.

Before I consider the above, I have a conceptual question about acc/dec of elevators. Consider two cases: the elevator moving downwards at constant acceleration a and the elevator moving downwards at constant deceleration a. Am I correct in saying the normal force acting on the person from the elevator is the same in both cases? I have done some maths and I appear to get the same result.

Onto the question: vel. of ball wrt earth = vel. of ball wrt hand + vel. of hand wrt earth = u + v(t), where u is given in the problem statement and since the elevator is at constant a, v = v(t). Denote vel. of ball wrt earth = $\dot{x}$. So $\dot{x} = u + v(t)\,\Rightarrow\, \dot{x} - v(t) = u(t)$. (I think the vel. of the ball wrt hand changes over time.) I can't solve this eqn since I have two functions dependent on t. I am now looking for some assumption to make on u(t) so I can solve for it first (i.e something like rate of change of u proportional to how fast u is intially or something like that).. Any ideas? (or am I mistaken somewhere else)

Many thanks.
EDIT: I just reliased I made v an arbritary function of t. It should be a linear function in t, so like v = pt. I don't think it solves my problem, however.

Last edited: Feb 20, 2013
2. Feb 20, 2013

### haruspex

No. A downward deceleration is an upward acceleration (even though the velocity be downwards).
In the question, what is the actual acceleration of the ball? What is the acceleration of the ball relative to the man's hand?

3. Feb 20, 2013

### voko

What forces act on the ball? What is its acceleration?

4. Feb 20, 2013

### CAF123

Yes, that makes sense which is why I am asking the question: When I do some maths, I get: consider constant deceleration downwards. Take y positive down: ma = mg -N => N = m(g-a). Where have I used the fact that we have dec/acc?
The only force acting on the ball is gravity: $\ddot{x} = -g$.

5. Feb 20, 2013

### voko

So you should be able to solve for u(t) then.

6. Feb 20, 2013

### Staff: Mentor

The simplest way to do this problem is to reference everything to the earth. You can solve for the time-dependent location of the ball relative to the earth and for the time-dependent location of the man's hand relative to the earth. Then you can subtract.

Let x0 represent the location of both the man's hand and the ball at time t = 0 when the ball is released. Also, let v0 represent the upward velocity of the man's hand when the ball is released. From the problem statement, in terms of v0 and u, what is the upward velocity of the ball relative to the earth when it is released? What is the man's acceleration relative to the earth? After the ball is released, what is the ball's acceleration relative to the earth? You now know the initial location, the initial velocity, and the acceleration both of the man's hand and of the ball relative to the earth. You can now solve for the time-dependent location of the ball relative to the earth and for the time-dependent location of the man's hand relative to the earth. Then you can subtract.

7. Feb 20, 2013

### CAF123

Is it just u(t) = -g(t - t1) + u? Now I should sub this into what I got for the diff eqn and solve:$$\ddot{x} - v(t) = -g(t - t_1) + u$$ But I still have t on the RHS?

8. Feb 20, 2013

### voko

$\ddot{x} - \dot{x}$ makes no sense physically.

9. Feb 20, 2013

### CAF123

I meant to write $\dot{x} - v(t) = -g(t-t_1) + u$

10. Feb 20, 2013

### voko

What is v(t)? Why?

11. Feb 20, 2013

### CAF123

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 =$\dot{u}$

12. Feb 20, 2013

### CAF123

v(t) is velocity of the elevator. Since the elevator is at a constant acceleration, $v \propto t$

13. Feb 20, 2013

### voko

What is v(t) EXACTLY?

14. Feb 20, 2013

### CAF123

Do you mean it's functional form? Assuming the elevator started at zero velocity, v(t) = at, where a is the acceleration of the lift.

15. Feb 20, 2013

### voko

given v(t) = at, how do you get $\dot{x} - v(t) = -g(t - t_1) + u$?

16. Feb 20, 2013

### CAF123

How did I derive it? I did everything in the earth frame:
(what we want to solve for is position of ball rel. to hand.)

vel. of ball rel. to hand = vel. of ball wrt earth - vel. of hand wrt earth, which means vel. of ball wrt earth = u(t) + at. Denote vel of ball wrt earth by $\dot{x}$. So then:
$\dot{x} - at = u(t)$ and I calculated u(t) earlier.

17. Feb 20, 2013

### voko

The problem is that u(t) = -g(t - t1) + u implies u'(t) = -g. So the acceleration of the ball with respect to the hand, which is accelerated with respect to the Earth, equals the acceleration of gravity. This cannot be correct.

18. Feb 20, 2013

### CAF123

But gravity is the only force acting on the ball?

19. Feb 20, 2013

### voko

In the inertial frame of the Earth - yes. What about the accelerated frame of the hand? That's where your u(t) is.

20. Feb 20, 2013

### CAF123

Once it has left the hand, then only gravity acts on the ball. But I suppose the initial push of the hand upwards would have given it some acceleration. So relative to the hand, we have an acceleration > g?