- #1
John O' Meara
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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
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