- #1

CAF123

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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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