Person throws ball in elevator problem

In summary: I meant to write ##\ddot{x} - v(t) = -g(t-t_1) + u##What is v(t)? Why?The velocity of the ball relative to the Earth at time t when it is released is v0.
  • #36
CAF123 said:
vel. of ball relative to earth: u + vo. Do I have to consider the velocity of the hand? The problem statement 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 [itex]x=x_0+v_0t+\frac{1}{2}at^2[/itex]?

Now all you have to do is write the corresponding equation for the location of the ball relative to the earth.
 
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  • #37
Chestermiller said:
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
 
  • #38
Yes. Sorry. I must have skipped over that previous post. Good job.

Chet
 
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