Person throws ball in elevator problem

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Homework Statement


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)

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.
 
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  • #2
haruspex
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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?
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
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What forces act on the ball? What is its acceleration?
 
  • #4
CAF123
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No. A downward deceleration is an upward acceleration (even though the velocity be downwards).

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?
In the question, what is the actual acceleration of the ball? What is the acceleration of the ball relative to the man's hand?

The only force acting on the ball is gravity: ##\ddot{x} = -g##.
 
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So you should be able to solve for u(t) then.
 
  • #6
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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
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So you should be able to solve for u(t) then.

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
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## \ddot{x} - \dot{x} ## makes no sense physically.
 
  • #9
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## \ddot{x} - \dot{x} ## makes no sense physically.

I meant to write ##\dot{x} - v(t) = -g(t-t_1) + u##
 
  • #10
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What is v(t)? Why?
 
  • #11
CAF123
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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?

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.
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?
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
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What is v(t)? Why?

v(t) is velocity of the elevator. Since the elevator is at a constant acceleration, ##v \propto t##
 
  • #13
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What is v(t) EXACTLY?
 
  • #14
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What is v(t) EXACTLY?
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
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given v(t) = at, how do you get ## \dot{x} - v(t) = -g(t - t_1) + u ##?
 
  • #16
CAF123
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given v(t) = at, how do you get ## \dot{x} - v(t) = -g(t - t_1) + u ##?

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
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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
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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.

But gravity is the only force acting on the ball?
 
  • #19
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In the inertial frame of the Earth - yes. What about the accelerated frame of the hand? That's where your u(t) is.
 
  • #20
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In the inertial frame of the Earth - yes. What about the accelerated frame of the hand? That's where your u(t) is.

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?
 
  • #21
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The hand ITSELF is accelerated with respect to the Earth. This is your accelerated reference frame. You should treat that as a non-inertial frame, with fictitious forces, or you should FIRST find the ball's velocity and the hand's velocity with respect to the Earth, and THEN subtract.
 
  • #22
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The hand ITSELF is accelerated with respect to the Earth. This is your accelerated reference frame. You should treat that as a non-inertial frame, with fictitious forces, or you should FIRST find the ball's velocity and the hand's velocity with respect to the Earth, and THEN subtract.

I see. I think I have it now: Let ##v_{XY}## = vel of X relative to Y So,
##v_{BH} = u - g(t - t_1) - at##.

Which means, distance of ball from hand relative to hand: ##\delta(s) = ut - \frac{1}{2}gt^2 - \frac{1}{2}at^2 + gt_1t##
Rather, ##\delta(s) = ut - ut_1 - \frac{1}{2}gt^2 - \frac{1}{2}gt_1^2 - \frac{1}{2}at^2 - \frac{1}{2}at_1^2 + gt_1t - gt_1^2## and then I can collect terms. This is nearly the answer in the other solution, but they don't have the last 2 terms.
 
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  • #23
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I see. I think I have it now: Let ##v_{XY}## = vel of X relative to Y So,
##v_{BH} = u - g(t - t_1) - at##.

Let ## t = t1 ##. Then ##v_{BH} ## must be ## u ##. But it is not.

Which means, distance of ball from hand relative to hand: ##\delta(s) ... ##

That should be ## v_{BH} ##. What is ##\delta(s) ## and what is ## s ## to begin with?
 
  • #24
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Let ## t = t1 ##. Then ##v_{BH} ## must be ## u ##. But it is not.

Stupid mistake. When I set up the expression I neglected that t1 wasn't 0. So it should read vBH = u -g(t-t1) -a(t-t1)


That should be ## v_{BH} ##. What is ##\delta(s) ## and what is ## s ## to begin with?
I am finding the displacement of the ball from the hand. We know the velocity so I integrated. In the end, I get $$x(t) - x(t1) = ut - ut_1 - 1/2 gt^2 + 1/2 gt_1^2 + gt_1t -gt_1^2 - 1/2 at^2 - 1/2 at_1^2 + at_1t -at_1^2$$
 
  • #25
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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.

What is the value of x at t1, by the way?
 

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