How High Was the Elevator When the Marble Was Dropped?

[John O' Meara]

An elevator ascends from the ground with uniform speed. At time T1 a marble is droped from the floor and hits the ground T2 seconds later. Find the height of the elevator at time T1? Clue T1=T2=4s.


Z0=V0z*T1

Z=V0z*T2 - (g*(T2)^2)/2.
Where Z0 is the height at T1, Z is taken to be vertical position, V0z is the initial speed in the Z direction (vertical).

Assuming that t=0 when the elevator passed the ground. The moment the marble leaves the elevator it initially continues moving in the same direction as the elevator but with reducing speed all the time until the marble stops momentary and then falls down at an ever increasing speed.
Z = Z0 +2*Za, where Za = coasting height of the marble between it leaving the elevator and its speed equaling zero. This migth be very simple I just don't see it. Please help. Thanks in advance

 

[ideasrule]

Z=V0z*T2 - (g*(T2)^2)/2.
Where Z0 is the height at T1, Z is taken to be vertical position, V0z is the initial speed in the Z direction (vertical).

Not quite. You have to add in the initial position of the marble, Z0=V0z*T1, to get the marble's position above the ground.

Assuming that t=0 when the elevator passed the ground. The moment the marble leaves the elevator it initially continues moving in the same direction as the elevator but with reducing speed all the time until the marble stops momentary and then falls down at an ever increasing speed.
Z = Z0 +2*Za, where Za = coasting height of the marble between it leaving the elevator and its speed equaling zero. This migth be very simple I just don't see it. Please help. Thanks in advance

You don't have to explicitly account for any of this. You have the equation:

Z=V0z*T1 + V0z*T2 - (g*(T2)^2)/2

z should be 0 because the marble reaches the ground at t=T2, and we've defined the ground to be z=0. So you can just solve for V0z, then multiply by T1 to get the marble's height at the moment of release.