Help with a mechanics problem from Kleppner's book

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The problem involves an elevator moving with uniform velocity and a marble dropped inside it, falling under gravity. The height at time T1 can be expressed using the equations h = vT1 and h = vT2 - 0.5g(T2)^2. A key insight is that when T1 equals T2, the height calculated should reflect the total distance traveled by both the elevator and the marble, which should sum to zero. The discussion highlights the need to check the signage in the equations to resolve discrepancies in height calculations. Properly accounting for the directions of motion is crucial for solving the problem accurately.
Tahmeed
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Homework Statement


An elevator departs the ground at t=0 With uniform velocity. At T1 a boy drops a marbel through the floor of the lift that falls under constant acceleration g=9.8 ms^-2. If it takes marbel T2 time to fall down, what was the height at the T1.

Homework Equations



h=vT1
h= vT2 -.5g(T2)^2

The Attempt at a Solution



I simply replaced v in second equation by h/T1. And i got the h in terms of T1, T2. But the problem is, there is a clue given with this problem in the book. That is when T1=T2=4 h=39.2 now, in my equation, T1=T2 gives a height 0.
 
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you have to relook your 2nd equation for proper signage.
Tahmeed said:

Homework Statement


An elevator departs the ground at t=0 With uniform velocity. At T1 a boy drops a marbel through the floor of the lift that falls under constant acceleration g=9.8 ms^-2. If it takes marbel T2 time to fall down, what was the height at the T1.

Homework Equations



h=vT1
h= vT2 -.5g(T2)^2

The Attempt at a Solution



I simply replaced v in second equation by h/T1. And i got the h in terms of T1, T2. But the problem is, there is a clue given with this problem in the book. That is when T1=T2=4 h=39.2 now, in my equation, T1=T2 gives a height 0.
the height that the elevator rises at T1 is the same as the height the marble falls at T2. The sum total of those 2 distances is 0 (h - h =0). Check signage.
 
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