Person throws ball in elevator problem

[Chestermiller]

vel. of ball relative to earth: u + v_o. 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 $x=x_0+v_0t+\frac{1}{2}at^2$?

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

 

[CAF123]

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]

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

Chet